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Concept 05 / 07Binomial Theorem

Numerically Greatest Term and Greatest Coefficient

Using the ratio of consecutive terms T(r+1)/T(r) to locate the numerically greatest term in (1 + x)^n, the case of equal greatest terms, the difference between greatest term and greatest coefficient, and extending the method to (a + b)^n.

Figure 1
The greatest binomial coefficient of (1+x)n(1 + x)^{n} is the middle one, (n⌊n/2⌋)\binom{n}{\lfloor n/2 \rfloor}, whatever xx is. The numerically greatest term is a different question, because each term also includes a power of xx, and that can shift the peak left or right. The ratio method. For (1+x)n(1 + x)^{n} with x>0x > 0, consecutive terms satisfy Tr+1Tr=(nr)xr(nr−1)xr−1=n−r+1rx\frac{T_{r+1}}{T_{r}} = \frac{\binom{n}{r}x^{r}}{\binom{n}{r-1}x^{r-1}} = \frac{n - r + 1}{r}x. The terms increase while this ratio exceeds 11. Solving (n−r+1)xr≥1\frac{(n - r + 1)x}{r} \ge 1 gives r≤(n+1)x1+xr \le \frac{(n + 1)x}{1 + x}. Let m=(n+1)x1+xm = \frac{(n + 1)x}{1 + x}. If mm is not an integer, the terms increase up to T⌊m⌋+1T_{\lfloor m \rfloor + 1} and then decrease, so the greatest term is T⌊m⌋+1T_{\lfloor m \rfloor + 1}. If mm is an integer, then Tm+1=TmT_{m+1} = T_{m} (the ratio at r=mr = m is exactly 11), and there are two equal greatest terms, TmT_{m} and Tm+1T_{m+1}. The figure shows n=10n = 10 and x=1.5x = 1.5: m=11×1.52.5=6.6m = \frac{11 \times 1.5}{2.5} = 6.6, so the terms rise up to T7T_{7} and fall afterwards. Note that the peak is to the right of the middle term T6T_{6}, because x>1x > 1 weights later terms more heavily. For x<1x < 1 the peak shifts left. Why this derivation is reliable: it checks directly when moving one step to the right increases the term, which is exactly what 'greatest' requires. The same method works for any sequence whose consecutive ratio is simple. Numerically greatest. For (1−x)n(1 - x)^{n} or negative xx, the signs alternate, and 'numerically greatest' means greatest absolute value. Since ∣Tr+1∣|T_{r+1}| is the same as in (1+∣x∣)n(1 + |x|)^{n}, apply the method with ∣x∣|x|. General binomial (a+b)n(a + b)^{n}. Factor out: (a+b)n=an(1+ba)n(a + b)^{n} = a^{n}(1 + \frac{b}{a})^{n}, and apply the method with x=∣ba∣x = |\frac{b}{a}|. Equivalently, use the ratio Tr+1Tr=n−r+1r⋅∣ba∣\frac{T_{r+1}}{T_{r}} = \frac{n - r + 1}{r} \cdot \left|\frac{b}{a}\right| directly. Greatest coefficient. If a question asks for the greatest coefficient of an expansion like (2+3x)n(2 + 3x)^{n}, the coefficient of xrx^{r} is (nr)2n−r3r\binom{n}{r}2^{n-r}3^{r}, which is again a 'term' problem, with xx replaced by 32\frac{3}{2}. Only for (1+x)n(1 + x)^{n} is the greatest coefficient simply the middle binomial coefficient. Read carefully whether the question asks for the greatest coefficient, the greatest binomial coefficient, or the greatest term at a specific value of xx. Key Takeaways: - Tr+1Tr=n−r+1rx\frac{T_{r+1}}{T_{r}} = \frac{n - r + 1}{r}x for (1+x)n(1 + x)^{n}. - m=(n+1)∣x∣1+∣x∣m = \frac{(n + 1)|x|}{1 + |x|}: greatest term T⌊m⌋+1T_{\lfloor m \rfloor + 1}; if mm is an integer, TmT_{m} and Tm+1T_{m+1} are equal greatest. - For (a+b)n(a + b)^{n} use x=∣ba∣x = |\frac{b}{a}|. - Greatest binomial coefficient is always the middle one; the greatest term depends on xx. - For x>1x > 1 the peak is right of the middle; for x<1x < 1 it is left. - Pitfall: answering with rr instead of the term number r+1r + 1. Brain-Teaser 1: Find the numerically greatest term in the expansion of (3−5x)15(3 - 5x)^{15} when x=15x = \frac{1}{5}. Solution: At x=15x = \frac{1}{5} the binomial is (3−1)15(3 - 1)^{15}, so ∣ba∣=13|\frac{b}{a}| = \frac{1}{3}. Then m=16×131+13=16/34/3=4m = \frac{16 \times \frac{1}{3}}{1 + \frac{1}{3}} = \frac{16/3}{4/3} = 4, an integer. So T4T_{4} and T5T_{5} are the equal numerically greatest terms. ∣T5∣=(154)311⋅14=1365×311|T_{5}| = \binom{15}{4}3^{11} \cdot 1^{4} = 1365 \times 3^{11}, and ∣T4∣=(153)312=455×312=1365×311|T_{4}| = \binom{15}{3}3^{12} = 455 \times 3^{12} = 1365 \times 3^{11}, confirming equality. The value ∣T4∣=∣T5∣=455×312|T_{4}| = |T_{5}| = 455 \times 3^{12}. Brain-Teaser 2: Find the greatest coefficient in the expansion of (1+2x)12(1 + 2x)^{12}. Solution: The coefficient of xrx^{r} is cr=(12r)2rc_{r} = \binom{12}{r}2^{r}, so crcr−1=13−rr×2≥1  ⟺  26−2r≥r  ⟺  r≤263≈8.67\frac{c_{r}}{c_{r-1}} = \frac{13 - r}{r} \times 2 \ge 1 \iff 26 - 2r \ge r \iff r \le \frac{26}{3} \approx 8.67. So the coefficients increase up to r=8r = 8 and then decrease, and the greatest is c8=(128)28=495×256=126720c_{8} = \binom{12}{8}2^{8} = 495 \times 256 = 126720. Compare with the greatest binomial coefficient (126)=924\binom{12}{6} = 924, which sits at r=6r = 6. The factor 2r2^{r} shifts the peak to the right. Brain-Teaser 3: For which positive xx is the 55th term the greatest in (1+x)10(1 + x)^{10}? Solution: We need T5≥T4T_{5} \ge T_{4} and T5≥T6T_{5} \ge T_{6}. T5T4=74x≥1\frac{T_{5}}{T_{4}} = \frac{7}{4}x \ge 1 gives x≥47x \ge \frac{4}{7}. T6T5=65x≤1\frac{T_{6}}{T_{5}} = \frac{6}{5}x \le 1 gives x≤56x \le \frac{5}{6}. So 47≤x≤56\frac{4}{7} \le x \le \frac{5}{6} (with ties at the endpoints; for strictly greatest, use strict inequalities). Reversing the method, fixing the position and solving for xx, is a common JEE variant.

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Numerically Greatest Term and Greatest Coefficient — DPP

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