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Beyond Advanced · Binomial Theorem

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1
Let nn be an integer with 2≤n≤1002\le n\le100. The expansion of (1+x)n(1+x)^n has a unique middle term, and three consecutive binomial coefficients, none outside the expansion, are in arithmetic progression. Find the sum of all possible values of nn.
Not attempted
A
208208
B
210210
C
416416
D
200200
2
Write (a+bx)n=∑r=0ncrxr(a+bx)^n=\sum_{r=0}^n c_rx^r, where a,b>0a,b>0 and n≥5n\ge5 is an integer. Suppose c3=3c2c_3=3c_2 and 10c4=21c310c_4=21c_3. The two middle coefficients are cm,cm+1c_m,c_{m+1}. Find cmcm+1/(cm−1cm+2)c_mc_{m+1}/(c_{m-1}c_{m+2}).
Not attempted
A
81/6481/64
B
25/1625/16
C
5/45/4
D
125/64125/64
3
For 2≤n≤2002\le n\le200, count the ordered pairs of integers (n,r)(n,r) with 1≤r<n/21\le r<n/2 such that the coefficients of xr−1,xr,xr+1x^{r-1},x^r,x^{r+1} in (1+x)n(1+x)^n are in arithmetic progression. Count each admissible pair once.
Not attempted
Your answer: —
Correct answer: 12
4
In the expansion of (2+3x)12(2+3x)^{12} let Tr+1=(12r)212−r(3x)rT_{r+1}=\binom{12}{r}2^{12-r}(3x)^r, and let x>0x>0 satisfy T5=T9T_5=T_9. Select all true statements.
Not attempted
A
x=2/3x=2/3
B
T7T_7 is uniquely greatest.
C
49T6T8=36T7249T_6T_8=36T_7^2
D
T1+T3+⋯+T13=224T_1+T_3+\cdots+T_{13}=2^{24}
5
The coefficients of x8,x9,x10x^8,x^9,x^{10} in (1+x)n(1+x)^n are in arithmetic progression, where n≥10n\ge10. The expansion has two middle terms. Find the difference between the binomial coefficient of either middle term and the immediately preceding coefficient on the increasing side.
Not attempted
A
13520781352078
B
429429
C
208012208012
D
416024416024
6
In (u−v)15(u-v)^{15}, where u,v>0u,v>0, terms are written in descending powers of uu. The sixth term from the beginning and the sixth term from the end have equal absolute values. Select all true statements.
Not attempted
A
u=vu=v
B
The two specified terms sum to zero.
C
Both middle terms are positive.
D
Exactly eight terms are positive.
7
Expand (2+23)90(\sqrt2+\sqrt[3]2)^{90} by the binomial theorem without collecting distinct terms. Let NN be the number of rational terms, and let jj be the term number of the middle rational term when the rational terms are ordered as in the full expansion. If there are two middle rational terms, use the earlier one. Find N+jN+j.
Not attempted
A
5858
B
6060
C
2424
D
5959
8
Let n≥2n\ge2 and a,b>0a,b>0. Write (a+bx)n=∑crxr(a+bx)^n=\sum c_rx^r. For 1≤r≤n−11\le r\le n-1 define Dr=cr2−cr−1cr+1D_r=c_r^2-c_{r-1}c_{r+1}. Select all statements valid for every such expansion.
Not attempted
A
Dr>0D_r>0
B
Dr/cr2=Dn−r/cn−r2D_r/c_r^2=D_{n-r}/c_{n-r}^2
C
The normalised difference is largest at a middle index.
D
No three consecutive coefficients form a geometric progression.
9
An expansion (1+x)n(1+x)^n, with even n≥4n\ge4, has its middle binomial coefficient equal to 12/712/7 times the coefficient two places before it. If SS is the sum of the coefficients of the three middle terms, find S/132S/132.
Not attempted
Your answer: —
Correct answer: 71.5
10
In the expansion of (2t+2−t)19(2^t+2^{-t})^{19}, let TjT_j denote the jjth term from the beginning. A real tt satisfies T7=16T14T_7=16T_{14}. Find T10/T11T_{10}/T_{11}.
Not attempted
A
24/72^{4/7}
B
22/72^{2/7}
C
28/72^{8/7}
D
21/72^{1/7}
11
For a real aa, the coefficient of x35x^{35} in Pa(x)=(1+x)18(1−x)16(1+x2)16(1+ax+x2)P_a(x)=(1+x)^{18}(1-x)^{16}(1+x^2)^{16}(1+ax+x^2) is zero. Let NN be the number of nonzero coefficients of the resulting polynomial, including its constant coefficient. Find N+a2N+a^2.
Not attempted
Your answer: —
Correct answer: 39
12
Find the constant term in (x2−2/x)9(x+1/x2)6(x^2-2/x)^9(x+1/x^2)^6, for x≠0x\ne0.
Not attempted
A
−1008-1008
B
10081008
C
504504
D
20162016
13
Evaluate ∑j=08(−1)j(8j)∑k=030(k12−2j)\displaystyle\sum_{j=0}^{8}(-1)^j\binom8j\sum_{k=0}^{30}\binom{k}{12-2j}, where a binomial coefficient is zero when its lower index is negative or exceeds its upper index.
Not attempted
A
4225662742256627
B
163996448163996448
C
−42256627-42256627
D
−248509702-248509702
14
In the uncollected binomial expansion of (23+34)84(\sqrt[3]2+\sqrt[4]3)^{84}, consider only the rational terms. Find the term number, in the full expansion, of the greatest rational term.
Not attempted
A
3737
B
4848
C
55
D
4949
15
Let aka_k be the coefficient of xkx^k in (1−x)2(1+x)20(1-x)^2(1+x)^{20}. Find a11/a10a_{11}/a_{10}.
Not attempted
A
4/34/3
B
3/43/4
C
2/32/3
D
−4/3-4/3
16
Let P(x)=(1+x3)8(1−x3)6(1+x6)P(x)=(1+x^3)^8(1-x^3)^6(1+x^6). Select all true statements.
Not attempted
A
The degree is 4848.
B
The coefficient of x25x^{25} is zero.
C
Exactly 1717 coefficients are nonzero.
D
The sum of all coefficients is zero.
17
Let P(x)=(1+x)9(1−x)7=∑r=016arxrP(x)=(1+x)^9(1-x)^7=\sum_{r=0}^{16}a_rx^r. Find ∑r=016ar2858\displaystyle\frac{\sum_{r=0}^{16}a_r^2}{858}.
Not attempted
Your answer: —
Correct answer: 17
18
After collection, let P(x)=(1+x+x2)5(1−x)5(1+x3)5P(x)=(1+x+x^2)^5(1-x)^5(1+x^3)^5. If SS is the sum of the squares of its nonzero coefficients and NN their number, find S/NS/N.
Not attempted
A
252252
B
4242
C
2121
D
3636
19
For 0<θ<π/20<\theta<\pi/2, in the expansion of (sin⁡θ+2cos⁡θ)10(\sin\theta+2\cos\theta)^{10} the fifth and seventh terms are equal. Find the ratio of the sixth term to the sum of all eleven terms.
Not attempted
A
63/12863/128
B
21/12821/128
C
63/25663/256
D
1/111/11
20
In the uncollected expansion of (53+74)77(\sqrt[3]5+\sqrt[4]7)^{77}, let j1<⋯<jNj_1<\cdots<j_N be the original term numbers of all rational terms. Find ∑k=1N(−1)k−1jk2\sum_{k=1}^N(-1)^{k-1}j_k^2. If the sum is negative, report its absolute value.
Not attempted
Your answer: —
Correct answer: 2808
21
For a positive integer nn, each of (2+3x)n(2+3x)^n and (3+2x)n(3+2x)^n has exactly two equal greatest coefficients. The smaller of the two maximising exponents in the first expansion exceeds the smaller maximising exponent in the second by 77. Find the sum of the original term numbers of these four greatest coefficients.
Not attempted
Your answer: —
Correct answer: 72
22
Let cr=(nr)an−rbrc_r=\binom nr a^{n-r}b^r, with a,b>0a,b>0, n≥4n\ge4. Suppose ck=ck+2c_k=c_{k+2} for some 0≤k≤n−20\le k\le n-2. Select all statements that must hold.
Not attempted
A
ck+1c_{k+1} is the unique greatest coefficient.
B
ck+12/ck2=(k+1)(n−k−1)(n−k)(k+2)c_{k+1}^2/c_k^2=\frac{(k+1)(n-k-1)}{(n-k)(k+2)}
C
nn must be even and k+1=n/2k+1=n/2.
D
ck+2>ckc_{k+2}>c_k
23
For x=k/100x=k/100 with integer 1≤k≤1001\le k\le100, the eighth term is the unique numerically greatest term in (1−x)20(1-x)^{20} and the fourth term is the unique greatest term in (1+x2)14(1+x^2)^{14}. Find the sum of all admissible kk.
Not attempted
Your answer: —
Correct answer: 555
24
For 1≤n≤1001\le n\le100, the same original term number is the unique greatest term in both (1+11/10)n(1+11/10)^n and (1+10/11)n(1+10/11)^n, each interpreted as a binomial expansion before addition. Find the sum of all such nn.
Not attempted
A
110110
B
180180
C
100100
D
9090
25
For t>0t>0, put ur=(20r)tru_r=\binom{20}{r}t^r for 0≤r≤200\le r\le20. Assume u8u_8 is the unique largest term. Select all correct statements about the second-largest value.
Not attempted
A
Only u7u_7 and u9u_9 can be second-largest.
B
If t<6/13t<\sqrt{6/13}, then u7u_7 is second-largest.
C
At t=6/13t=\sqrt{6/13} there are two equal greatest terms.
D
If t>6/13t>\sqrt{6/13}, then u9u_9 is second-largest.
26
Consider the twenty-one signed terms of the binomial expansion of (2−3)20(2-3)^{20}, before adding them. Let MM be the largest signed term and mm the smallest signed term. Find M/∣m∣M/|m|.
Not attempted
A
13/1213/12
B
12/1312/13
C
26/2726/27
D
27/2627/26
27
For 0≤r≤120\le r\le12 define vr=2r(12r)(1812−r)v_r=2^r\binom{12}{r}\binom{18}{12-r}. Let kk be its unique maximising index. Find 13vk+1/vk−113v_{k+1}/v_{k-1}, correct to two decimal places.
Not attempted
Your answer: —
Correct answer: 14
28
Let P(x)=(1+x)15P(x)=(1+x)^{15} and Q(x)=P(x)+xP′(x)Q(x)=P(x)+xP\prime(x). After substituting x=3/5x=3/5 into the individual monomials of QQ, let jj be the original term number of the greatest monomial value. Find jj.
Not attempted
A
55
B
77
C
88
D
66
29
For 0≤r≤150\le r\le15, let vr=(15r)/((r+1)(r+2))v_r=\binom{15}{r}/((r+1)(r+2)). Select all true statements.
Not attempted
A
Exactly v6,v7v_6,v_7 are greatest.
B
v8=v9v_8=v_9
C
∑r=015vr=(217−18)/272\sum_{r=0}^{15}v_r=(2^{17}-18)/272
D
vr=v15−rv_r=v_{15-r} for every rr.
30
For fixed n≥2n\ge2, put pr(t)=(nr)tr/(1+t)np_r(t)=\binom nr t^r/(1+t)^n for t>0t>0. Select all statements valid for every such nn.
Not attempted
A
At most two indices can maximise pr(t)p_r(t).
B
The maximum height is unchanged by replacing tt with 1/t1/t.
C
Exactly nn positive values of tt give a tied greatest pair.
D
The maximum height is always minimised at t=1t=1.
31
Let mm be a positive integer and define S=∑r=02m(−1)r(2mr)2S=\sum_{r=0}^{2m}(-1)^r\binom{2m}{r}^2 and W=∑r=02m(−1)r(r−m)2(2mr)2W=\sum_{r=0}^{2m}(-1)^r(r-m)^2\binom{2m}{r}^2. Select all identities valid for every mm.
Not attempted
A
S=(−1)m(2mm)S=(-1)^m\binom{2m}{m}
B
W=−m2SW=-m^2S
C
(−1)m+1W>0(-1)^{m+1}W>0 for every mm.
D
S+W/m2=0S+W/m^2=0
32
For an integer n≥2n\ge2, define A=∑r=0n(nr)2A=\sum_{r=0}^n\binom nr^2 and B=∑r=0n(nr)2/(r+1)B=\sum_{r=0}^n\binom nr^2/(r+1). Find B/AB/A.
Not attempted
A
2n(n+1)2\frac{2n}{(n+1)^2}
B
1n+1\frac1{n+1}
C
2n+1(n+1)2\frac{2n+1}{(n+1)^2}
D
n+12n+1\frac{n+1}{2n+1}
33
Evaluate ∑r=010(−1)r(r−3)2(10r)(20−r10)\displaystyle\sum_{r=0}^{10}(-1)^r(r-3)^2\binom{10}{r}\binom{20-r}{10}.
Not attempted
Your answer: —
Correct answer: 4559
34
For integers n≥k≥1n\ge k\ge1, put A=∑r=0n−k(nr)(nr+k)A=\sum_{r=0}^{n-k}\binom nr\binom n{r+k} and B=∑r=0n−kr(nr)(nr+k)B=\sum_{r=0}^{n-k}r\binom nr\binom n{r+k}. Select all identities valid for every such pair.
Not attempted
A
A=(2nn−k)A=\binom{2n}{n-k}
B
2B=(n−k)A2B=(n-k)A
C
B=nA/2B=nA/2
D
B=n(2n−1n−k−1)B=n\binom{2n-1}{n-k-1}
35
Let S=∑0≤r≤17, 3∣rr(r−1)(17r)S=\sum_{0\le r\le17,\ 3\mid r}r(r-1)\binom{17}{r}. Find 3S/(17⋅16)−2153S/(17\cdot16)-2^{15}.
Not attempted
Your answer: —
Correct answer: 1
36
For an integer n≥2n\ge2, evaluate ∑r=0n(2r−n)2(nr)2(2nn)\displaystyle\frac{\sum_{r=0}^n(2r-n)^2\binom nr^2}{\binom{2n}{n}}.
Not attempted
A
nn
B
n/2n/2
C
n2/(2n+1)n^2/(2n+1)
D
n2/(2n−1)n^2/(2n-1)
37
For n≥1n\ge1 an integer, let S=∑r=0n(−1)r(r−2)2(nr)(r+2)(r+3)S=\sum_{r=0}^n\frac{(-1)^r(r-2)^2\binom nr}{(r+2)(r+3)}. Find (n+1)(n+2)(n+3)S(n+1)(n+2)(n+3)S.
Not attempted
A
16n−216n-2
B
16n+216n+2
C
16n+2316n+23
D
16n−5016n-50
38
For 0≤r≤n0\le r\le n, let Ar=∑k=0r(−2)k(rk)A_r=\sum_{k=0}^r(-2)^k\binom rk and Br=∑k=0rk(−2)k(rk)B_r=\sum_{k=0}^rk(-2)^k\binom rk, where n≥2n\ge2. Select all correct identities.
Not attempted
A
∑r=0n(nr)Ar2=2n\sum_{r=0}^n\binom nr A_r^2=2^n
B
∑r=0n(nr)ArBr=n2n\sum_{r=0}^n\binom nr A_rB_r=n2^n
C
∑r=0n(nr)Br2=n22n\sum_{r=0}^n\binom nr B_r^2=n^22^n
D
Br=2rArB_r=2rA_r for all rr.
39
Twelve labelled positions each carry one of A,B,C,DA,B,C,D. Exactly four positions carrying AA are additionally marked; unmarked positions may also carry AA. Require at least one unmarked AA and at least one BB. Let NN count all symbol assignments and choices of marks. Find N/330N/330.
Not attempted
Your answer: —
Correct answer: 79005
40
Match each expression in List-I to its value in List-II. Here Cr=(8r)C_r=\binom8r and a coefficient with lower index outside 0,…,80,\ldots,8 is zero. List-IList-IIP) 9∑r=08(−1)rCrr+11) 1Q) 9∑r=08Crr+12) 70R) ∑r=06CrCr+23) 511S) ∑r=08(−1)rCr24) 80085) 12870\begin{array}{ll}\text{List-I}&\text{List-II}\\ \text{P)}\ 9\sum_{r=0}^8\frac{(-1)^rC_r}{r+1}&\text{1)}\ 1\\ \text{Q)}\ 9\sum_{r=0}^8\frac{C_r}{r+1}&\text{2)}\ 70\\ \text{R)}\ \sum_{r=0}^6 C_rC_{r+2}&\text{3)}\ 511\\ \text{S)}\ \sum_{r=0}^8(-1)^rC_r^2&\text{4)}\ 8008\\ &\text{5)}\ 12870\end{array}
Not attempted
A
P-1, Q-2, R-4, S-3
B
P-1, Q-3, R-4, S-2
C
P-3, Q-1, R-5, S-2
D
P-2, Q-3, R-1, S-4
41
Let P(x)=(1+x)12(1+x+x2)4=∑arxrP(x)=(1+x)^{12}(1+x+x^2)^4=\sum a_rx^r. Find ∑3∣r(r−10)2ar/210\displaystyle\sum_{3\mid r}(r-10)^2a_r/2^{10}.
Not attempted
Your answer: —
Correct answer: 612
42
For an integer n≥2n\ge2, suppose ∑r=0n(2r−n)2(nr)2=14423∑r=0n(nr)2\sum_{r=0}^n(2r-n)^2\binom nr^2=\frac{144}{23}\sum_{r=0}^n\binom nr^2. Find the number of terms with positive coefficient in (1−x)n+(1+x)n(1-x)^n+(1+x)^n.
Not attempted
Your answer: —
Correct answer: 7
43
The number N=(624+424−2)/50N=(6^{24}+4^{24}-2)/50 is an integer. Find its least nonnegative remainder on division by 125125.
Not attempted
A
7676
B
2626
C
5151
D
11
44
Find the largest integer kk such that 2k2^k divides 3192−13^{192}-1.
Not attempted
Your answer: —
Correct answer: 8
45
Find the remainder when ⌊(2+3)25⌋\lfloor(2+\sqrt3)^{25}\rfloor is divided by 10001000.
Not attempted
A
201201
B
124124
C
125125
D
123123
46
Let A=2+5A=2+\sqrt5, Ij=⌊Aj⌋I_j=\lfloor A^j\rfloor, and fj=Aj−Ijf_j=A^j-I_j. Select all true statements.
Not attempted
A
I17I_{17} is even.
B
I18I_{18} is odd.
C
A35f17(1−f18)=1A^{35}f_{17}(1-f_{18})=1
D
f17+f18<1f_{17}+f_{18}<1
47
Let A=3+22A=3+2\sqrt2. Find the last two digits, interpreted as a number from 00 to 9999, of ⌊A20+A21⌋\lfloor A^{20}+A^{21}\rfloor.
Not attempted
Your answer: —
Correct answer: 71
48
Let X=(7+6)20X=(\sqrt7+\sqrt6)^{20}. Find the remainder when ⌊X⌋\lfloor X\rfloor is divided by 100100.
Not attempted
A
7373
B
7474
C
2525
D
9999
49
Let ff be the fractional part of (1+2)13(1+\sqrt2)^{13}. Find ⌊1/f⌋\lfloor1/f\rfloor.
Not attempted
A
9464194641
B
9464294642
C
4732147321
D
9464394643
50
For each integer n≥3n\ge3, put An=(1+1/n)nA_n=(1+1/n)^n and Bn=(1+1/n)n+1B_n=(1+1/n)^{n+1}. Select all statements true for every such nn.
Not attempted
A
2<An<32<A_n<3
B
Bn<4B_n<4
C
(n+1)n<nn+1(n+1)^n<n^{n+1}
D
Bn<3B_n<3
51
For integers n≥2n\ge2, define En=9n−8n−1E_n=9^n-8n-1. Select all true statements.
Not attempted
A
The greatest common divisor of all EnE_n is 6464.
B
128128 divides every EnE_n.
C
En/64≡(n2)(mod8)E_n/64\equiv\binom n2\pmod8
D
If 16∣n16\mid n, then 512∣En512\mid E_n.
52
Let N=10131−1−3100−(312)104106N=\frac{101^{31}-1-3100-\binom{31}{2}10^4}{10^6}. Find the remainder when the integer NN is divided by 10410^4.
Not attempted
A
10951095
B
44954495
C
995995
D
90059005
53
In (x−2y+4z)12(x-2y+4z)^{12} consider only monomials whose exponents of xx and zz are equal. Let MM be the greatest coefficient among these monomials. Find M/212M/2^{12}.
Not attempted
Your answer: —
Correct answer: 34650
54
Let NN be the number of distinct monomials in (u+v+w)8(u+v+w)^8, where u,v,wu,v,w are independent variables. Substitute u=1,v=x,w=x3u=1,v=x,w=x^3 and collect powers of xx; let MM be the number of nonzero coefficients. Find N−MN-M.
Not attempted
A
4545
B
2020
C
2424
D
2121
55
For a real aa, the coefficient of x2x^2 in (1+ax−x2)10(1+ax-x^2)^{10} is zero. If cc is the coefficient of x4x^4 in that same polynomial, find −27c-27c.
Not attempted
Your answer: —
Correct answer: 665
56
Expand (x+y+z)12(x+y+z)^{12} into distinct monomials. Let EE be the sum of coefficients of monomials whose three exponents are even, and let NN be the number of those monomials. Select all true statements.
Not attempted
A
N=28N=28
B
E=(312+3)/4E=(3^{12}+3)/4
C
No monomial has exactly one odd exponent.
D
The coefficient of x4y4z4x^4y^4z^4 equals E/NE/N.
57
In (2x−y+z)10(2x-y+z)^{10} sum the coefficients of monomials in which the exponents of x,zx,z are both positive and the exponent of yy is odd. Find this signed sum.
Not attempted
A
−493740-493740
B
493740493740
C
−523776-523776
D
−494252-494252
58
In (u+v+w+z)18(u+v+w+z)^{18} retain precisely the monomials whose exponents of uu and vv are equal. Let NN be the number of retained distinct monomials and SS the sum of their coefficients. Which ordered pair is (N,S)(N,S)?
Not attempted
A
(100,(213))(100,\binom{21}{3})
B
(100,(3618))(100,\binom{36}{18})
C
(190,(3618))(190,\binom{36}{18})
D
(55,418)(55,4^{18})
59
Twelve independent labelled positions each carry one of A,B,CA,B,C, with all 3123^{12} assignments equally likely. Let EE mean that A,BA,B occur equally often and that the positive number of occurrences of CC is divisible by four. If pp is the probability of EE, find 312p3^{12}p.
Not attempted
Your answer: —
Correct answer: 37621
60
In the collected expansion of (x−y+z)17(x-y+z)^{17}, let MM be the largest absolute value of a coefficient. Select all true statements.
Not attempted
A
Exactly three coefficients have magnitude MM.
B
Exactly two coefficients equal MM.
C
Exactly one coefficient equals −M-M.
D
A monomial with exponent pattern 7,5,57,5,5 can also have magnitude MM.
61
In the uncollected multinomial expansion of (2+3+6)15(\sqrt2+\sqrt3+\sqrt6)^{15}, let NN be the number of rational terms, with one term for each exponent triple. Among these terms let MM be the largest multinomial coefficient, excluding the radical-power factor. Find M/NM/N.
Not attempted
Your answer: —
Correct answer: 27027
62
Let P(x,y,z)=(x+y+z)12−(x+y−z)12P(x,y,z)=(x+y+z)^{12}-(x+y-z)^{12}. Select all correct statements after like monomials have been collected.
Not attempted
A
Exactly 4242 distinct monomials survive.
B
No surviving monomial has equal exponents of xx and yy.
C
The sum of coefficients is 2⋅3112\cdot3^{11}.
D
The coefficient of x5y6zx^5y^6z is 1108811088.
63
For an integer n≥3n\ge3, the quotient Q(x)=((1+x)n−1−nx)/x2Q(x)=((1+x)^n-1-nx)/x^2 is extended polynomially to x=0x=0. Its coefficient of xx is five times its constant coefficient. Select all true statements.
Not attempted
A
deg⁡Q=15\deg Q=15
B
Its greatest coefficients are at x6,x7x^6,x^7.
C
The sum of its even-power coefficients is 6553665536.
D
The alternating sum of its coefficients is 1616.
64
Evaluate lim⁡x→0243x4((1+x)1/3−1−x3+x29−5x381)\displaystyle\lim_{x\to0}\frac{243}{x^4}\left((1+x)^{1/3}-1-\frac{x}{3}+\frac{x^2}{9}-\frac{5x^3}{81}\right), where the real cube root is used. A solution using an infinite series is not required or assumed.
Not attempted
A
−5-5
B
1010
C
−10-10
D
00
65
Find the absolute value of lim⁡x→064x4(1+x+1−x−2+x24)\displaystyle\lim_{x\to0}\frac{64}{x^4}\left(\sqrt{1+x}+\sqrt{1-x}-2+\frac{x^2}{4}\right). Use real square roots.
Not attempted
Your answer: —
Correct answer: 5
66
Real numbers a,ba,b satisfy lim⁡x→0(1+ax+bx2)4−(1+x)7x2=0\displaystyle\lim_{x\to0}\frac{(1+ax+bx^2)^4-(1+x)^7}{x^2}=0. Find lim⁡x→032((1+ax+bx2)4−(1+x)7)x3\displaystyle\lim_{x\to0}\frac{32\big((1+ax+bx^2)^4-(1+x)^7\big)}{x^3}.
Not attempted
Your answer: —
Correct answer: 7
67
For x≠1x\ne1 set Q(x)=((1−x13)/(1−x))4Q(x)=((1-x^{13})/(1-x))^4, and extend it polynomially at x=1x=1. Find the coefficient of x26x^{26} in QQ.
Not attempted
Your answer: —
Correct answer: 1420
68
Find ⌊108[(101100)20−1−15−191000]⌋\displaystyle\left\lfloor10^8\left[\left(\frac{101}{100}\right)^{20}-1-\frac15-\frac{19}{1000}\right]\right\rfloor.
Not attempted
Your answer: —
Correct answer: 119003
69
For 0<x<10<x<1, let yy be the positive solution of y3=1+xy^3=1+x, and put L=1+x/3−x2/9L=1+x/3-x^2/9, U=L+5x3/81U=L+5x^3/81. Select all statements valid for every such xx.
Not attempted
A
L<y<UL<y<U
B
U<1+x/3U<1+x/3
C
0<y−L<5x3/810<y-L<5x^3/81
D
L>yL>y
70
Find ⌊1012(100110003−1−13000+19000000)⌋\displaystyle\left\lfloor10^{12}\left(\sqrt[3]{\frac{1001}{1000}}-1-\frac1{3000}+\frac1{9000000}\right)\right\rfloor.
Not attempted
Your answer: —
Correct answer: 61
A coarse decimal approximation of the cube root is insufficient. Turn the two candidate integer bounds into exact rational cube comparisons.
Put b=1+1/3000−1/9000000b=1+1/3000-1/9000000 and y=1001/10003y=\sqrt[3]{1001/1000}. We need the integer part of 1012(y−b)10^{12}(y-b). Take l=b+61/1012l=b+61/10^{12} and u=b+62/1012u=b+62/10^{12}. Both are positive rationals. Expanding the finite cubes, or clearing their denominators, gives l3<1001/1000<u3l^3<1001/1000<u^3. For a transparent check, b3−(1+1/1000)=−(1/1000)3729(135−9/106+1/109)b^3-(1+1/1000)=-\frac{(1/1000)^3}{729}(135-9/10^6+1/10^9); then add 3b2d+3bd2+d33b^2d+3bd^2+d^3 with d=61/1012d=61/10^{12} or 62/101262/10^{12} to obtain the respective signs. Since cubing is strictly increasing on positive numbers, 61<1012(y−b)<6261<10^{12}(y-b)<62. The required floor is 6161. Rounding the root before subtracting bb loses the small residual.