Question

Expand to 4 terms the following expressions :
$(1+x^2{)}^{-2}$

Answer 1

Whatever be value of $n$, negative integer or positive or negative fraction,

$(1+x{)}^n=1+\frac{n}{1!}x+\frac{n(n-1)}{2!}{x}^2+\frac{n(n-1)(n-2)}{3!}{x}^3+...+...+\frac{n(n-1)(n-2)...(n-r+1)}{r!}{x}^r+...+$ to $\infty$

where, $|x|<1$ i.e. $-1<x<1.$

Now,

$(1+x^2{)}^{-2}=1-2x^2+\frac{-2(-2-1)}{1.2}(x^2{)}^2+\frac{-2(-2-1)(-2-2)}{\mathrm{1.2.3}}(x^2{)}^3+...........$

$=1-2x^2+3x^4-4x^6+.......$