Question

How do you prove that: `1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+n) = (2n)/(n+1)` ?

Answer 1

True

Explanation:

Supposing that the afirmation is true

`1+1/(1+2)+1/(1+2+3)+1/(1+2+3+4)+cdots+1/(1+2+3+4+cdots+n)= sum_(k=1)^n 2/(k(k+1)) = S_n`

`S_1 = (2 xx 1)/(1+1)=1`

so supposing is true

`S_n = (2n)/(n+1)` then

`S_(n+1) = S_n + 2/((n+1)(n+2)) = (2(n+1))/(n+2)` which is the expected result.

Then by mathematical induction, it is true the assertion.

Answer 2

Prove by induction.

Explanation:

Proof by induction:

Base case

If `n = 1` then then left hand side is:

`1`

and the right hand side is:

`(2(1))/((1)+1) = 2/2 = 1`

So the equation holds for `n=1`

Induction step

First note that:

`1+2+3+...+n = 1/2 n(n+1)`

Suppose:

`1+1/(1+2)+1/(1+2+3)+...+1/(1+2+3+...+n) = (2n)/(n+1)`

Then:

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

`= (2n)/(n+1)+1/(1+2+3+...+n+(n+1))`

`= (2n)/(n+1)+1/(1/2 (n+1)(n+2))`

`= (2n)/(n+1)+2/((n+1)(n+2))`

`= ((2n)(n+2))/((n+1)(n+2))+2/((n+1)(n+2))`

`= (2n^2+4n+2)/((n+1)(n+2))`

`= (2(n^2+2n+1))/((n+1)(n+2))`

`= (2(n+1)^2)/((n+1)(n+2))`

`= (2(n+1))/(n+2)`

Answer 3

`" The Sum="(2n)/(n+1).`

Explanation:

We will use a Method called Method of Telescopic Sum.

Denoting by, `T_j`, the `j^(th)` term of the Series, we observe that,

`T_j=1/(1+2+...+j)=1/(sumj)=1/(j/2(j+1))=2/(j(j+1).`

`:." The Reqd. Sum, say S"=sum_(j=1)^(j=n)T_j=sum_(j=1)^(j=n)2/(j(j+1))`

`=2sum_(j=1)^(j=n)1/(j(j+1))=2sum_(j=1)^(j=n){(j+1)-j}/(j(j+1))`

`=2sum_(j=1)^(j=n){(j+1)/(j(j+1))-j/(j(j+1))}`

`=2sum_(j=1)^(j=n){1/j-1/(j+1)}`

`=2[(1/1-cancel(1/2))+(cancel(1/2)-cancel(1/3))+(cancel(1/3)-cancel(1/4))+...+(cancel(1/(n-1))-cancel(1/n))+(cancel(1/n)-1/(n+1))]`

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

`:." The Sum="(2n)/(n+1)`, as Respected George C., and,

Respected Cesareo R. , have derived!

Enjoy Maths.!