Sequences & SerieseasyFree

Sum of (2^k - 1)/k! to Infinity | JEE

JEE Maths question with a full step-by-step solution.

Question
The value of the series is ∑k=1∞1k!(∑n=1k2n−1)\displaystyle\sum_{k=1}^{\infty}\frac{1}{k!}\left(\sum_{n=1}^{k}2^{n-1}\right)
Aee
Be2+ee^2+e
Ce2e^2
De2−ee^2-ecorrect
Solution
Step 1: The inner sum is 1+2+22+⋯+2k−1=2k−1.1+2+2^2+\cdots+2^{k-1}=2^k-1. Step 2: So
∑k=1∞2k−1k!=∑k=1∞2kk!−∑k=1∞1k!=(e2−1)−(e−1)=e2−e.\sum_{k=1}^{\infty}\frac{2^k-1}{k!}=\sum_{k=1}^{\infty}\frac{2^k}{k!}-\sum_{k=1}^{\infty}\frac{1}{k!}=(e^2-1)-(e-1)=e^2-e.
 Correct answer: (4)
Still stuck on this question?Ask your doubt on WhatsApp
Similar questions
Sequences & Series · easy
a, b, c, d are in increasing G.P. If the A.M. between a and b is 6 and the A.M. between c…
Sequences & Series · medium
Let a₁, a₂, …, a(4001) be in A.P. such that (1)/(a₁ a₂)+(1)/(a₂ a₃)+⋯+ 1 a(4000) a(4001)…
Sequences & Series · medium
If (p+q· 5^x)/(p-q· 5^x)=(q+r· 5^x)/(q-r· 5^x)=(r+s· 5^x)/(r-s· 5^x) , then p, q, r, s ar…
Sequences & Series · easy
Let a₁, a₂, … and b₁, b₂, … be arithmetic progressions with a₁=25 , b₁=75 and a(100)+b(10…

Solve more, learn faster

Sign up free to solve more JEE Maths questions and explore doMath — timed drills, mastery sprints, bookmarks, and chapter-wise progress tracking.

Sign up freeExplore doMath