First slide

Binomial theorem for positive integral Index

Question

If for some positive integer $n$ the coefficients of three consecutive terms in the binomial expansion of ${(1 + x)}^{n + 5}$ are in the ratio $5 : 10; 14, .$ then the largest coefficient- in the expansion is

Moderate

A

330
B

252
C

792
D

462

Solution

Let three consecutive terms be $T_{r}, T_{r + 1}$ and $T_{r + 2}$
Their coefficients are $^{n + 5} C_{r - 1},^{n + 5} C_{r}$ and $^{n + 5} C_{r} + 1$ .It is given that
$^{n + 5} C_{r - 1} :^{n + 5} C_{r} :^{n + 5} C_{r + 1} = 5 : 10 : 14$
$\frac{^{n + 5} C_{r}}{^{n + 5} C_{r - 1}} = \frac{10}{5}$ and $\frac{^{n + 5} C_{r + 1}}{^{n + 5} C_{r}} = \frac{14}{10}$
$\Rightarrow \frac{n + 5 - r + 1}{r} = 2$ and $\frac{n + 5 - r}{r + 1} = \frac{7}{5} [∵ \frac{n_{r}}{n_{r - 1}} = \frac{n - r + 1}{r}]$
$\Rightarrow n - r + 6 = 2 r$ and $5 (n - r + 5) = 7 r + 7$
$\Rightarrow n - 3 r + 6 = 0$ and $5 n - 12 r + 18 = 0$
$\Rightarrow n = 6, r = 4$
Thus, the coefficients in the binomial expansion of $(1 + x)^{11}$ are
$^{11} C_{0},^{11} C_{1}, \dots,^{11} C_{11}$ .The greatest coefficient is $^{11} C_{5} =^{11} C_{6} = 462$

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