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7.Binomial Theorem
hard
If ${a_r}$ is the coefficient of ${x^r}$, in the expansion of ${(1 + x + {x^2})^n}$, then ${a_1} - 2{a_2} + 3{a_3} - .... - 2n\,{a_{2n}} = $
A
$0$
B
$n$
C
$-n$
D
$2n$
Solution
(c) Let us take ${a_0} + {a_1}x + {a_2}{x^2} + ….. + {a_{2n}}{x^{2n}} = {(1 + x + {x^2})^n}$
Differentiating with respect to x on both sides ${a_1} + 2{a_2}x + … + 2n\,{a_{2n}}{x^{2n – 1}}$ = $n{(1 + x + {x^2})^{n – 1}}(2x + 1)$
Put $x = -1$ ==> ${a_1} – 2{a_2} + 3{a_3} – …. + 2n\,{a_{2n}} = – n$.
Standard 11
Mathematics