For some n neq 10,let the coefficients of the | vedclass

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(B) The coefficients of the $5^{	ext{th}}$,$6^{	ext{th}}$,and $7^{	ext{th}}$ terms in $(1+x)^{n+4}$ are $^{n+4}C_4$,$^{n+4}C_5$,and $^{n+4}C_6$ res

Vedclass · Accepted Answer

$35$

Vedclass · Answer

(B) The coefficients of the $5^{	ext{th}}$,$6^{	ext{th}}$,and $7^{	ext{th}}$ terms in $(1+x)^{n+4}$ are $^{n+4}C_4$,$^{n+4}C_5$,and $^{n+4}C_6$ respectively.Since they are in $A.P.$,we have $2 	imes ^{n+4}C_5 = ^{n+4}C_4 + ^{n+4}C_6$.Dividing by $^{n+4}C_5$,we get $2 = \frac{^{n+4}C_4}{^{n+4}C_5} + \frac{^{n+4}C_6}{^{n+4}C_5}$.Using the formula $\frac{^{n}C_r}{^{n}C_{r-1}} = \frac{n-r+1}{r}$,we get $2 = \frac{5}{n+4-4+1} + \frac{n+4-6+1}{6} = \frac{5}{n+1} + \frac{n-1}{6}$.$12(n+1) = 30 + (n-1)(n+1)$ $\Rightarrow 12n + 12 = 30 + n^2 - 1$ $\Rightarrow n^2 - 12n + 17 = 0$ (Wait,let us re-evaluate).Using $2 	imes ^{n+4}C_5 = ^{n+4}C_4 + ^{n+4}C_6$:$2 	imes \frac{(n+4)!}{5!(n-1)!} = \frac{(n+4)!}{4!n!} + \frac{(n+4)!}{6!(n-2)!}$.Dividing by $(n+4)!$ and multiplying by $6!n!$: $2 	imes 6n = 6 	imes 5 	imes n + (n)(n-1)$ $\Rightarrow 12n = 30n + n^2 - n$ $\Rightarrow n^2 + 17n = 0$ (Incorrect approach).Correct expansion: $2 	imes \frac{(n+4)!}{5!(n-1)!} = \frac{(n+4)!}{4!n!} + \frac{(n+4)!}{6!(n-2)!}$.Divide by $(n+4)!$ and multiply by $6!n!$: $2 	imes 6n = 6 	imes 5 + n(n-1)$ $\Rightarrow 12n = 30 + n^2 - n$ $\Rightarrow n^2 - 13n + 30 = 0$.$(n-10)(n-3) = 0$. Since $n 
eq 10$,$n = 3$.Then $n+4 = 7$. The expansion is $(1+x)^7$.The largest coefficient is the middle term coefficient,$^{7}C_3 = \frac{7 	imes 6 	imes 5}{3 	imes 2 	imes 1} = 35$.

For some $n \neq 10$,let the coefficients of the $5^{\text{th}}$,$6^{\text{th}}$,and $7^{\text{th}}$ 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:

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