Question ID: #375
For some $n \ne 10$, let the coefficients of the $5^{th}$, $6^{th}$ and $7^{th}$ terms in the binomial expansion of $(1+x)^{n+4}$ be in A.P. Then the largest coefficient in the expansion of $(1+x)^{n+4}$ is:
- (1) 70
- (2) 35
- (3) 20
- (4) 10
Solution:
Let $N = n+4$. The coefficients are $^NC_4$, $^NC_5$, $^NC_6$.
Since they are in A.P.:
$$ 2(^NC_5) = ^NC_4 + ^NC_6 $$
Divide by $^NC_5$:
$$ 2 = \frac{^NC_4}{^NC_5} + \frac{^NC_6}{^NC_5} $$
Using property $\frac{^nC_r}{^nC_{r-1}} = \frac{n-r+1}{r}$:
$$ 2 = \frac{5}{N-4} + \frac{N-5}{6} $$
$$ 12(N-4) = 30 + (N-5)(N-4) $$
$$ 12N – 48 = 30 + N^2 – 9N + 20 $$
$$ N^2 – 21N + 98 = 0 $$
$$ (N-7)(N-14) = 0 $$
So $N=7$ or $N=14$.
Given $n \ne 10 \Rightarrow n+4 \ne 14 \Rightarrow N \ne 14$.
Therefore, $N = 7$.
We need the largest coefficient in $(1+x)^7$.
For odd power $N=7$, the largest coefficients are the middle terms: $^7C_3$ and $^7C_4$.
$$ ^7C_3 = \frac{7 \times 6 \times 5}{3 \times 2 \times 1} = 35 $$
Ans. (2)
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