The sum to n terms of the series 1 + [2(1 + 1/n)] + [3(1+1/n)^2] +.... is?
Please help me with the above question.
Thank You!
Sanyam are you sure that this sequence is a series......................its not forming an A.P. as well as a G.P.
Hi Deepshikha, actually i know this does not form either AP or GP. But there are plenty other methods with which such series summation is done but I am not able to find that method.
Perhaps you are not aware.
Perhaps you are not aware.
It seems to evaluate to n^2, on solving for n=1, 2, 3, 4, and 5. I didn't go beyond that as you have to solve for the power of 5 and so on.
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It is of the form of an AGP with d=1 and r=(1+1/n), but I was not able to solve it using the general formula... perhaps some error in simplification.
It is a series, of the form:
Hope this helps!
I solved the problem and the answer is indeed n^2. It is an arithmetico-geometric progression, and I have started with the general form of the sum and then solved it for this case:
(AGPs are of the form: a, (a+d)r, (a+2d)r^2,...)
Hope you could follow the solution, and that it helped!
Hey Milind,
Thank You very much.
Thank You very much.
Hi Friend,
S = 1 + [2(1+1/n)] + [3(1+1/n)^2] + .........
S(1+1/n) = + (1+1/n) 2(1+1/n)^2 3(1+1/n).............
-S(1/n) = 1 + (1+1/n) + (1+1/n)^2.............
-S(1/n) = 1/1 - (1 +1/n) = -n
so S = n^2 ans.
S = 1 + [2(1+1/n)] + [3(1+1/n)^2] + .........
S(1+1/n) = + (1+1/n) 2(1+1/n)^2 3(1+1/n).............
-S(1/n) = 1 + (1+1/n) + (1+1/n)^2.............
-S(1/n) = 1/1 - (1 +1/n) = -n
so S = n^2 ans.