The sum to n terms of the series 1+2 (1+ 1 n )+3 (1+ 1 n )^ 2 + . is given by
Mathematics · JEE Main · NTA Exams — Progression and Series
The sum to n terms of the series
\(1+2\left(1+\frac{1}{n}\right)+3\left(1+\frac{1}{n}\right)^{2}+\ldots .\) is given by
\(1+2\left(1+\frac{1}{n}\right)+3\left(1+\frac{1}{n}\right)^{2}+\ldots .\) is given by
- n2
- n (n + 1)
- n (1 + 1/n)2
- none of these
Answer
(A) n 2
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- The sum to n terms of the series 1 1.3 + 1 3.5 + 1 5.7 + ..... is
- Let T r be the r th term of an A.P., for r = 1, 2, 3,....... If for some positive integers m, n. We have T_ m…
- n arithmetic means are inserted in between x and 2y and then between 2x and y. In case the r th mean in both…
- The sum of the infinite series is :
- For what value of m, a ^ m+1 + b ^ m+1 a ^ m + b ^ m is the arithmetic mean of 'a' and 'b' ?
- If the first and the nth terms of a G.P. are a and b respectively and P is the product of the first n terms…
- Let S_ k = _ r=1 ^ k ^ -1 ( 6^ r 2^ 2 r+1 +3^ 2 r+1 ) Then _ k S_ k is equal to
- If x , y , z are in A.P. and tan –1 x , tan –1 y and tan –1 z are also in A.P., then
More Progression and Series questions · Browse all practice questions