Sequence And Series Question 353
Question: $ 1+\frac{2^{4}}{2,!}+\frac{3^{4}}{3,!}+\frac{4^{4}}{4,!}+…..\infty = $
Options:
A) $ 5,e $
B) $ e $
C) $ 15,e $
D) $ 2,e $
Show Answer
Answer:
Correct Answer: C
Solution:
$ 1+\frac{2^{4}}{2!}+\frac{3^{4}}{3!}+\frac{4^{4}}{4!}+….\infty $ $ T_{n}=\frac{n^{4}}{n!}=\frac{n^{3}}{(n-1)!}=\frac{n^{3}-1}{(n-1)!}+\frac{1}{(n-1)!} $ $ =\frac{(n-1)(n^{2}+n+1)}{(n-1)!}+\frac{1}{(n-1)!}=\frac{n^{2}+n+1}{(n-2)!}+\frac{1}{(n-1)!} $ $ =\frac{n^{2}-4}{(n-2)!}+\frac{(n-2)}{(n-2)!}+\frac{7}{(n-2)!}+\frac{1}{(n-1)!} $ $ =\frac{n+2}{(n-3)!}+\frac{1}{(n-3)!}+\frac{7}{(n-2)!}+\frac{1}{(n-1)!} $ = $ \frac{1}{(n-4)!}+\frac{6}{(n-3)!}+\frac{7}{(n-2)!}+\frac{1}{(n-1)!} $ = $ ,e+6e+7e+e=15e $ .
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