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જો ${(1 + x)^n}$ ના વિસ્તરણમાં ${p^{th}}$, ${(p + 1)^{th}}$ અને ${(p + 2)^{th}}$ પદો સમાંતર શ્રેણીમાં હોય તો . . . .
${n^2} - 2np + 4{p^2} = 0$
${n^2} - n\,(4p + 1) + 4{p^2} - 2 = 0$
${n^2} - n\,(4p + 1) + 4{p^2} = 0$
એકપણ નહીં.
Solution
(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 $(b).$
