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

AP ICET 2017 Shift 1 Paper

Question : 130 of 200

Marks: +1, -0

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

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}

....(i)
For

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

...(ii)

(i) — (ii) gives

2(a_{1}+a_{3}+a_{5}+. . .+a_{2n-1})=3^{n}-1

⇒a_{1}+a_{3}+a_{5}+. . .+a_{2n-1}={1}/{2}(3^{n}-1)

∑↙{k=1}↖{n}a_{2k-1}={1}/{2}(3^{n}-1)

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