Expand using Binomial Theorem (1+x2−2x)4,x≠0 and let the sum of coefficients of the terms in the expansion be t. Find 10000t
⇒(1+x2−2x)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)4−8x(1+x2)3+24x2(1+x2)2−32x3(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+x22−4x−5+16x+8x2−32x3+16x4
This is the final binomial expansion of above expression
∴sum of coefficients=116+12+12−4−5+16+8−32+16=116=t
⇒10000t=625