Question

Question #80860

Answer 1

Notice that the given be written as follows

`(1/1-1/2)+(1/2-1/3)+(1/3-1/4)+...+(1/n-1/(n+1))`

regroup and we have

`(1/1)+(-1/2+1/2)+(-1/3+1/3)+(-1/n+1/n)-1/(n+1)`

The sums inside the parentheses are zero hence we left with

`(1/1-1/((n+1)))=(n+1-1)/(n+1)=n/(n+1)`

Answer 2

See below.

Explanation:

For `n=1` is true for

`1/(1 xx 2) = 1/2`

Supposing that is true for `n`

`sum_(k=1)^n1/(k(k+1)) = n/(n+1)`

then it must be true for `n+1`

`sum_(k=1)^(n+1)1/(k(k+1)) = sum_(k=1)^n1/(k(k+1)) + 1/((n+1)(n+2))`
`= n/(n+1)+ 1/((n+1)(n+2)) = (n(n+2))/((n+1)(n+2))+ 1/((n+1)(n+2))`
`=(n+1)^2/((n+1)(n+2)) = (n+1)/(n+2)`

then by mathematical induction, the affirmation is true.