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Concept 04 / 07Binomial Theorem

Binomial Theorem for Positive Integral Index: General Term and Middle Terms

Statement and counting proof of (x + a)^n, the general term T(r+1) = nCr x^(n-r) a^r, symmetry of coefficients, middle terms for even and odd n, and the expansion of (x - a)^n and (1 + x)^n.

Figure 1
For a positive integer nn, (x+a)n=∑r=0n(nr)xn−rar=(n0)xn+(n1)xn−1a+(n2)xn−2a2+⋯+(nn)an(x + a)^{n} = \sum_{r=0}^{n}\binom{n}{r}x^{n-r}a^{r} = \binom{n}{0}x^{n} + \binom{n}{1}x^{n-1}a + \binom{n}{2}x^{n-2}a^{2} + \cdots + \binom{n}{n}a^{n}. Why the coefficients are binomial coefficients. Write (x+a)n(x + a)^{n} as a product of nn brackets (x+a)(x+a)⋯(x+a)(x + a)(x + a)\cdots(x + a). Expanding means choosing one term from each bracket and multiplying. A product equal to xn−rarx^{n-r}a^{r} arises exactly when aa is chosen from rr of the brackets and xx from the rest, and there are (nr)\binom{n}{r} ways to decide which rr brackets supply aa. This counting view explains every property below and extends immediately to multinomial expansions. (An induction proof uses Pascal's rule (nr)+(nr−1)=(n+1r)\binom{n}{r} + \binom{n}{r-1} = \binom{n+1}{r}.) Basic features. There are n+1n + 1 terms. In each term the powers of xx and aa add up to nn. The coefficients read the same from both ends, (nr)=(nn−r)\binom{n}{r} = \binom{n}{n - r}, as the figure shows for n=8n = 8. Replacing aa by −a-a gives (x−a)n=∑r(−1)r(nr)xn−rar(x - a)^{n} = \sum_{r}(-1)^{r}\binom{n}{r}x^{n-r}a^{r}, with alternating signs. With x=1x = 1: (1+a)n=∑(nr)ar(1 + a)^{n} = \sum\binom{n}{r}a^{r}. General term. The (r+1)(r + 1)th term from the beginning is Tr+1=(nr)xn−rarT_{r+1} = \binom{n}{r}x^{n-r}a^{r}. Note the off-by-one: T1T_{1} corresponds to r=0r = 0. This formula is the key to almost every question: to find a particular term, write down Tr+1T_{r+1}, simplify the powers of the variable, and solve for rr. The (r+1)(r + 1)th term from the end is the (n−r+1)(n - r + 1)th term from the beginning, or equivalently the (r+1)(r + 1)th term of the expansion of (a+x)n(a + x)^{n}. Middle terms. If nn is even, there is one middle term, Tn/2+1T_{n/2 + 1}. If nn is odd, there are two, T(n+1)/2T_{(n+1)/2} and T(n+3)/2T_{(n+3)/2}. The middle binomial coefficient is the largest: (nn/2)\binom{n}{n/2} for even nn, and the equal pair (n(n−1)/2)=(n(n+1)/2)\binom{n}{(n-1)/2} = \binom{n}{(n+1)/2} for odd nn. Combining expansions. (x+a)n+(x−a)n=2[(n0)xn+(n2)xn−2a2+⋯ ](x + a)^{n} + (x - a)^{n} = 2[\binom{n}{0}x^{n} + \binom{n}{2}x^{n-2}a^{2} + \cdots] keeps only the terms with even powers of aa, and (x+a)n−(x−a)n(x + a)^{n} - (x - a)^{n} keeps only the odd ones. This is the standard way to simplify expressions such as (2+1)6+(2−1)6(\sqrt{2} + 1)^{6} + (\sqrt{2} - 1)^{6}, where the irrational parts cancel. Key Takeaways: - (x+a)n=∑r=0n(nr)xn−rar(x + a)^{n} = \sum_{r=0}^{n}\binom{n}{r}x^{n-r}a^{r}; n+1n + 1 terms; exponents add to nn. - Tr+1=(nr)xn−rarT_{r+1} = \binom{n}{r}x^{n-r}a^{r}: always start a term question from this. - (x−a)n(x - a)^{n} has alternating signs: Tr+1=(−1)r(nr)xn−rarT_{r+1} = (-1)^{r}\binom{n}{r}x^{n-r}a^{r}. - Middle term: Tn/2+1T_{n/2+1} for even nn; T(n+1)/2T_{(n+1)/2} and T(n+3)/2T_{(n+3)/2} for odd nn. - (r+1)(r + 1)th term from the end =(n−r+1)= (n - r + 1)th from the start. - (x+a)n+(x−a)n(x + a)^{n} + (x - a)^{n} keeps even-position terms (even powers of aa) and doubles them. - Pitfall: using Tr=(nr)xn−rarT_{r} = \binom{n}{r}x^{n-r}a^{r} for the rrth term. That formula gives the (r+1)(r + 1)th term. Brain-Teaser 1: Find the middle term of (x3+3x)10\left(\frac{x}{3} + \frac{3}{x}\right)^{10}. Does its value depend on xx? Solution: n=10n = 10 is even, so the middle term is T6T_{6}, with r=5r = 5: T6=(105)(x3)5(3x)5=252T_{6} = \binom{10}{5}\left(\frac{x}{3}\right)^{5}\left(\frac{3}{x}\right)^{5} = 252. The powers of xx and of 33 cancel exactly, so the middle term is the constant 252252, independent of xx. This happens whenever the two terms of the binomial are reciprocals of each other and nn is even: the middle term is the constant term. Brain-Teaser 2: Simplify (3+1)5−(3−1)5(\sqrt{3} + 1)^{5} - (\sqrt{3} - 1)^{5} and use it to show that (3+1)5(\sqrt{3} + 1)^{5} is not an integer. Solution: The difference keeps twice the terms with odd powers of 11, that is, r=1,3,5r = 1, 3, 5: 2[(51)(3)4+(53)(3)2+(55)]=2[5×9+10×3+1]=2×76=1522[\binom{5}{1}(\sqrt{3})^{4} + \binom{5}{3}(\sqrt{3})^{2} + \binom{5}{5}] = 2[5 \times 9 + 10 \times 3 + 1] = 2 \times 76 = 152. So (3+1)5=152+(3−1)5(\sqrt{3} + 1)^{5} = 152 + (\sqrt{3} - 1)^{5}. Since 0<3−1<10 < \sqrt{3} - 1 < 1, the extra term lies strictly between 00 and 11, so (3+1)5(\sqrt{3} + 1)^{5} lies strictly between 152152 and 153153 and is not an integer. Its integer part is 152152. Pairing an expression with its conjugate is the standard tool for integer and fractional parts. Brain-Teaser 3: In the expansion of (1+x)n(1 + x)^{n}, the coefficients of the 55th, 66th and 77th terms are in arithmetic progression. Find nn. Solution: The coefficients are (n4)\binom{n}{4}, (n5)\binom{n}{5} and (n6)\binom{n}{6}, so 2(n5)=(n4)+(n6)2\binom{n}{5} = \binom{n}{4} + \binom{n}{6}. Divide by (n5)\binom{n}{5} and use the ratio (nr)(nr−1)=n−r+1r\frac{\binom{n}{r}}{\binom{n}{r-1}} = \frac{n - r + 1}{r}: (n4)(n5)=5n−4\frac{\binom{n}{4}}{\binom{n}{5}} = \frac{5}{n - 4} and (n6)(n5)=n−56\frac{\binom{n}{6}}{\binom{n}{5}} = \frac{n - 5}{6}. So 2=5n−4+n−562 = \frac{5}{n - 4} + \frac{n - 5}{6}, giving 12(n−4)=30+(n−5)(n−4)12(n - 4) = 30 + (n - 5)(n - 4), which simplifies to n2−21n+98=0n^{2} - 21n + 98 = 0. So n=7n = 7 or n=14n = 14. Both work: for n=7n = 7 the coefficients are 35,21,735, 21, 7, and for n=14n = 14 they are 1001,2002,30031001, 2002, 3003. Converting to ratios of consecutive coefficients avoids messy factorials.

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Binomial Theorem for Positive Integral Index: General Term and Middle Terms — DPP

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