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1/(1+x) =â(n: 0->â) (-1)^n .x^n
1/(1-x) = â(n: 0->â) x^n
consider
f(x) = 1/(1+x) =>f(0) = 1
f'(x) =-1/(1+x)^2 => f'(0)/1! = -1
...
...
f^(n)(x) = (-1)^n .n!/(1+x)^(n+1) => f^(n)(0)/n! = (-1)^n
1/(1+x)
=f(x)
=f(0) +[f'(0)/1!]x ++[f''(0)/2!]x^2+....+[f^(n)(0)/n!]x^n +...
=â(n: 0->â) (-1)^n .x^n
1/(1-x) = â(n: 0->â) x^n