mathematical induction, prove the expression with summation

Question

how to prove by mathematical induction the below expression:

can someone help me by proving it by the standart way

Answer 1

Suppose the basis step holds. So the sum you've written is assumed to be true. Then check the inductive step.

$$\frac{1}{(n+1)(n+2)} +\sum_{k}^{n} \frac{1}{k(k+1)} = \frac{n}{n+1} + \frac{1}{(n+1)(n+2)}$$

Then simplify the right hand side,

$$ = \frac{n(n+2)+1}{(n+1)(n+2)} = \frac{(n+1)^2}{(n+1)(n+2)}=\frac{n+1}{n+2}.$$

You can put it together from there? Establish the basis step, then polish up the inductive step above.

Answer 2

let * represent your function

-let n=1 then ${\sum^{n}_1}\frac{1}{k(k+1)}=\frac{1}{1(2)}=\frac{1}{1+1}$

-Assume * holds at n=k

-prove * holds for n=k+1

$\frac{k}{k+1}+\frac{1}{(k+1)(k+2)}= \frac{k+1}{k+2}$

notice here we just use our assumption and substitute $\frac{k}{k+1}$ into our LHS to represent the sum of values from 1 to k. Then we just add our one new value where n=k+1. We use n=k+1 on our RHS and now we will show the equality holds.

$\frac{k(k+2)+1}{(k+1)(k+2)}$

$\frac{(k+1)^2}{(k+1)(K+2)}$

$\frac{k+1}{k+2}=\frac{k+1}{k+2}$

since * holds for k+1, * holds for all n within the naturals by induction.