If coefficients of 2^{nd}, 3^{rd} and 4^{th} | vedclass

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Question

(d) Coefficients of $2^{nd}$, $3^{rd}$ and $4^{th}$ terms are respectively $^n{C_1},{\,^n}{C_2}$ and $^n{C_3}$ are in $A.P.$

==> ${2.^n}{C_2} =

Vedclass · Accepted Answer

$-14$

Vedclass · Answer

(d) Coefficients of $2^{nd}$, $3^{rd}$ and $4^{th}$ terms are respectively $^n{C_1},{\,^n}{C_2}$ and $^n{C_3}$ are in $A.P.$

==> ${2.^n}{C_2} = {\,^n}{C_1} + {\,^n}{C_3}$ 

==> $\frac{{2n!}}{{2\,!\left( {n - 2} \right)\,!}} $

= $\frac{{n!}}{{(n - 1)!}} + \frac{{n!}}{{3!\,\left( {n - 3} \right)\,!}}$ On solving, ${n^2} - 9n + 14 = 0\,\,$

$\Rightarrow \,\,{n^2} - 9n = - 14$.

If coefficients of $2^{nd}$, $3^{rd}$ and $4^{th}$ terms in the binomial expansion of ${(1 + x)^n}$ are in $A.P.$, then ${n^2} - 9n$ is equal to

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