Solveeit Logo
Login

Question

Mathematics Question on Sequence and series

The value of 23!+45!+67!+........\frac{2}{3!}+\frac{4}{5!}+\frac{6}{7!}+........ is

A

e12e^{\frac{1}{2}}

B

e−1e^{-1}

C

ee

D

e13e^{\frac{1}{3}}

Answer

e−1e^{-1}

Explanation

Solution

Given, 23!+45!+67!+…∞\frac{2}{3 !}+\frac{4}{5 !}+\frac{6}{7 !}+\ldots \infty
The nnth term of the above series
Tn=2n(2n+1)!=(2n+1)−1(2n+1)!=2n+1(2n+1)!−1(2n+1)!T_{n}=\frac{2 n}{(2 n+1) !}=\frac{(2 n+1)-1}{(2 n+1) !}=\frac{2 n+1}{(2 n+1) !}-\frac{1}{(2 n+1) !}
Tn=1(2n)!−1(2n+1)!T_{n}=\frac{1}{(2 n) !}-\frac{1}{(2 n+1) !}
∴T1=12!−13!\therefore T_{1}=\frac{1}{2 !}-\frac{1}{3 !}
T2=14!−15!T_{2}=\frac{1}{4 !}-\frac{1}{5 !}
T3=16!−17!T_{3}=\frac{1}{6 !}-\frac{1}{7 !}
... ... ...\begin{matrix}.&.&.\\\ .&.&.\\\ .&.&.\end{matrix}
Hence, Tn=12!−13!+14!−15!+…∞=e−1T_{n}=\frac{1}{2 !}-\frac{1}{3 !}+\frac{1}{4 !}-\frac{1}{5 !}+\ldots \infty=e^{-1}