A) 3e
B) 5e
C) 6e
D) 7e
Correct Answer: B
Solution :
[b] \[\because 1+\frac{{{2}^{3}}}{2}+\frac{{{3}^{3}}}{3}+\frac{{{4}^{3}}}{4}+..........to\,\,\infty \] \[\therefore {{t}_{n}}=\frac{{{n}^{3}}}{n}=\frac{n.{{n}^{2}}}{n\,\,n-1}=\frac{{{n}^{2}}}{n-1}\] \[=\frac{{{n}^{2}}+1-1}{n-1}=\frac{{{n}^{2}}-1}{n-1}+\frac{1}{n-1}=\frac{\left( n+1 \right)\left( n-1 \right)}{n-1}+\frac{1}{n-1}\] \[=\frac{n+1}{n-1}+\frac{1}{n-1}=\frac{n-2+2+1}{n-1}+\frac{1}{n-1}=\frac{1}{n-3}+\frac{3}{n-2}+\frac{1}{n-1}\] \[\therefore {{S}_{n}}=\sum\limits_{n=1}^{\infty }{{{t}_{n}}}=\sum\limits_{n=1}^{\infty }{\left[ \frac{1}{n-3}+\frac{3}{n-2}+\frac{1}{n-1} \right]}\] \[=\sum\limits_{n=1}^{\infty }{\frac{1}{n-3}}+\sum\limits_{n=1}^{\infty }{\frac{1}{n-2}+\sum\limits_{n=1}^{\infty }{\frac{1}{n-1}}}\] \[0=e+3e+e=5e\] Hence, the option [b] is correct.
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