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Question

Expand using Binomial Theorem (1+x22x)4,x0 and let the sum of coefficients of the terms in the expansion be t. Find 10000t

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Solution

(1+x22x)4

Let a=1+x2,b=2x

(40)(1+x2)4(41)(1+x2)3(2x)+(42)(1+x2)2(2x)2(43)(1+x2)1(2x)3+(44)(2x)4

(1+x2)48x(1+x2)3+24x2(1+x2)232x3(1+x2)1+(2x)4

We will use binomial expansion for (1+x2)4,(1+x2)3

[1+4.x2+6.x24+4.x38+x416]8x[1+3.x2+3x24+x38]+24x2[1+x24+x]32x3[1+x2]+16x4

=x416+x32+x224x5+16x+8x232x3+16x4

This is the final binomial expansion of above expression

sum of coefficients=116+12+1245+16+832+16=116=t

10000t=625

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