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If ( 1 − X + X 2 ) N = a 0 + a 1 X + a 2 X 2 + . . . + a 2 N X 2 N , Find the Value of a 0 + a 2 + a 4 + . . . + a 2 N . - Mathematics

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Question

If  \[\left( 1 - x + x^2 \right)^n = a_0 + a_1 x + a_2 x^2 + . . . + a_{2n} x^{2n}\] , find the value of  \[a_0 + a_2 + a_4 + . . . + a_{2n}\] .

 

Solution

Putting x = 1 and −1 in \[\left( 1 - x + x^2 \right)^n = a_0 + a_1 x + a_2 x^2 + . . . + a_{2n} x^{2n}\] we get,

\[1 = a_0 + a_1 + a_2 + . . . + a_{2n} . . . (1)\]

and

\[3^n = a_0 - a_1 + a_2 - . . . + a_{2n} . . . (2)\]

Adding (1) and (2), we get

\[3^n + 1 = 2\left( a_0 + a_2 + . . . + a_{2n} \right)\]
Hence, the value of  \[a_0 + a_2 + a_4 + . . . + a_{2n}\]  is \[\frac{3^n + 1}{2}\] .
 

 

 

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Proof of Binomial Therom by Combination
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Chapter 18: Binomial Theorem - Exercise 18.3 [Page 46]

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RD Sharma Mathematics [English] Class 11
Chapter 18 Binomial Theorem
Exercise 18.3 | Q 18 | Page 46
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