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

3 Answers
Sep 26, 2016

Answer:

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.

Sep 26, 2016

Answer:

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)#

Sep 26, 2016

Answer:

#" 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.!