A) 6 e
B) 6 e ? 1
C) 5 e
D) 5 e + 1
Correct Answer: B
Solution :
Let \[S=4+11+22+37+....+{{T}_{n-1}}+{{T}_{n}}\] or S = 4 + 11 + 22 + 37+??..+\[{{T}_{n-1}}+{{T}_{n}}\] \[\therefore \] On subtracting we get \[0=4+[7+11+15+19+....+({{T}_{n}}-{{T}_{n-1}})]-{{T}_{n}}\] \[0=4+\frac{n-1}{2}[14+(n-2)4]-{{T}_{n}}\] \[\therefore {{T}_{n}}=2{{n}^{2}}+n+1\] Thus \[n{{}^{th}}\]term of given series is \[{{T}_{n}}=\frac{2{{n}^{2}}+n+1}{(n)!}=\frac{2n}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}\] \[=\frac{2(n-1+1)}{(n-1)!}+\frac{1}{(n-1)!}+\frac{1}{n!}\]\[=\frac{2}{(n-2)!}+\frac{3}{(n-1)!}+\frac{1}{n!}\] \[\therefore \] Sum \[=\sum\limits_{n=1}^{\infty }{{{T}_{n}}=2e+3e+e-1=6e-1}\].
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