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Binomial Theorem — Maths STD 11 Science — Question

CBSE BoardEnglish MediumSTD 11 ScienceMathsBinomial Theorem1 Mark
Question
If $(1-\text{x}+\text{x}^{2})^{\text{n}}=\text{a}^{0}+\text{a}_{1}\text{x}+\text{a}_{2}\text{x}^{2}+...+\text{a}_{2\text{n}}\text{x}^{2\text{n}},$find the value of $\text{a}_{0}+\text{a}_{2}+\text{a}_{4}+...+\text{a}_{2\text{n}}.$
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Answer

Putting x = 1 and -1 in $(1-\text{x}+\text{x}^{2})^{\text{n}}=\text{a}^{0}+\text{a}_{1}\text{x}+\text{a}_{2}\text{x}^{2}+...+\text{a}_{2\text{n}}\text{x}^{2\text{n}}$
we get,
$1=\text{a}_{0}+\text{a}_{1}+\text{a}_{2}+....+\text{a}_{2\text{n}}\ ...(\text{1})$ and,
$3^{\text{n}}=\text{a}_{0}-\text{a}_{1}+\text{a}_{2}-....+\text{a}_{2\text{n}}\ ...(\text{2})$
Adding (i) and (ii), we get
$3^{\text{n}}+1=2(\text{a}_{0}+\text{a}_{2}+...+\text{a}_{2\text{n}})$
Hence, the value of $\text{a}_{0}+\text{a}_{2}+\text{a}_{4}+..\text{a}_{2\text{n}}$ is $\frac{3^{\text{n}}+1}{2}.$

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