Question

Question #5fced

Answer 1

`sum_(n=0)^oo(-1)^n/((2n+1)!)(pi/6)^(2n+1)=sin(pi/6)=1/2`

Explanation:

The Maclaurin series for `sin(x)` is given by

`sin(x)=sum_(n=0)^oo(-1)^n/((2n+1)!)x^(2n+1)`

Thus,

`sum_(n=0)^oo(-1)^n/((2n+1)!)(pi/6)^(2n+1)=sin(pi/6)=1/2`