If the coefficients of {p^{th}}, {(p + 1)^{th | vedclass

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Question

(b) Coefficient of ${p^{th}},{(p + 1)^{th}}$

and ${(p + 2)^{th}}$ terms in expansion of ${(1 + x)^n}$ are $^n{C_{p - 1}}{,^n}{C_p}{,^n}{C_{p + 1}}$

Vedclass · Accepted Answer

${n^2} - n\,(4p + 1) + 4{p^2} - 2 = 0$

Vedclass · Answer

(b) Coefficient of ${p^{th}},{(p + 1)^{th}}$

and ${(p + 2)^{th}}$ terms in expansion of ${(1 + x)^n}$ are $^n{C_{p - 1}}{,^n}{C_p}{,^n}{C_{p + 1}}$. 

Then ${2^n}{C_p} = {\,^n}{C_{p - 1}} + {\,^n}{C_{p + 1}}$

$ \Rightarrow {n^2} - n(4p + 1) + 4{p^2} - 2 = 0$

$Trick :$ Let $p = 1$ , hence $^n{C_0},{\,^n}{C_1}$and $^n{C_2}$are in $ A.P.$ 

$ \Rightarrow \,\,\,{2.^n}{C_1} = {\,^n}{C_0}{ + ^n}{C_2}\,\,\,\,$

$\Rightarrow 2n = 1 + \frac{{n\,(n - 1)}}{2}$$ \Rightarrow \,\,4n = 2 + {n^2} - n\,\,\,\,$ 

$\Rightarrow {n^2} - 5n + 2 = 0$ which is given by $(b).$

If the coefficients of ${p^{th}}$, ${(p + 1)^{th}}$ and ${(p + 2)^{th}}$ terms in the expansion of ${(1 + x)^n}$ are in $A.P.$, then

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