The coefficient of `x^20` in the expansion of `(1+x^2)^40.(x^2+2+1/x^2

1.	The coefficient of `x^20` in the expansion of `(1+x^2)^40.(x^2+2+1/x^2)^-5` is:
Answer» `(1+x^2)^40(x^2+2+1/x^2)^(-5)` `=(1+x^2)^40[(x+1/x)^2]^-5` `=(1+x^2)^40[((x^2+1)/x)^2]^-5` `=(1+x^2)^30[1/x^2]^-5` `=(1+x^2)^30x^10` `=x^10(C(30,0)*1+C(30,1)x^2+C(30,2)x^4+C(30,3)x^6+C(30,4)x^8+C(30,5)x^10+...C(30,30)x^60)` So, term with involving `x^20` will be `C(30,5)x^20`. So, coefficient of `x^20` will be `C(30,5).` Also, `C(30,5) = C(30,25)` So, option `(b)` is the correct option.

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The coefficient of `x^20` in the expansion of `(1+x^2)^40.(x^2+2+1/x^2)^-5` is:

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