(1+x+x^{2})^{n}=\sum_{k=0}^{2n}a_{k}x^{k}\Rightarrow\sum_{k=0}^{n}a_{2k}

AP ICET 2017 Shift 2 Paper

Question : 130 of 200

Marks: +1, -0

(1+x+x^{2})^{n}=∑_{k=0}^{2n}a_{k}x^{k}⇒∑_{k=0}^{n}a_{2k}

Solution:

Given :

∑↙{k=0}↖{2n}a_{k}x^{k}=a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+. . .+a_{2n}x^{2n}

=(1+x+x^2)^n

For

x=1:a_{0}+a_{1}+a_{2}+a_{3}+a_{4}+. . .+a_{2n}=3^{n}

For

x=–1:a_{0}-a_{1}+a_{2}-a_{3}+a_{4}+. . .+a_{2n}=1

or,

∑↙{k=0}↖{n}a_{2k}=\;\;{1}/{2}(3^{n}+1)

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