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Question

Expand to 4 terms the following expression :
(1+x)12

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Solution

Whatever be value of n, negative integer or positive or negative fraction,
(1+x)n=1+n1!x+n(n1)2!x2+n(n1)(n2)3!x3+...+...+n(n1)(n2)...(nr+1)r!xr+...+to
where, |x|<1 i.e. 1<x<1.
Now,
(1+x)12=1+12x+12(121)1.2x2+12(121)(122)1.2.3x3+.........
=1+12x18x2+116x3+.....

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