Sequence And Series Question 106
Question: The sum of 20 terms of the series whose rth term is given by $ T(n)={{(-1)}^{n}}\frac{n^{2}+n+1}{n!} $ is
Options:
A) $ \frac{20}{19!}-2 $
B) $ \frac{21}{20!}-1 $
C) $ \frac{21}{20!} $
D) none of these
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ T_{r}={{(-1)}^{r}}\frac{r^{2}+r+1}{r!} $ $ ={{(-1)}^{r}}[ \frac{r}{(r-1)!}+\frac{1}{(r-1)!}+\frac{1}{r!} ] $ $ ={{(-1)}^{r}}[ \frac{1}{(r-2)!}+\frac{1}{(r-1)!}+\frac{1}{(r-1)!}+\frac{1}{r!} ] $ $ =[ \frac{{{(-1)}^{r}}}{r!}+\frac{{{(-1)}^{r}}}{(r-1)!} ]+[ \frac{{{(-1)}^{r}}}{(r-1)!}+\frac{{{(-1)}^{r}}}{(r-2)!} ] $ $ =[ \frac{{{(-1)}^{r}}}{r!}-\frac{{{(-1)}^{r-1}}}{(r-1)!} ]-[ \frac{{{(-1)}^{r-1}}}{(r-1)!}-\frac{{{(-1)}^{r-2}}}{(r-2)!} ] $ $ =V(r)-V(e-1) $
$ \therefore \sum\limits_{r=1}^{n}{T_{r}}=V(n)-V(0)=[ \frac{{{(-1)}^{n}}}{n!}-\frac{{{(-1)}^{n-1}}}{(n-1)!} ]-1 $ Therefore, the sum of 20 terms is $ [ \frac{1}{20!}-\frac{-1}{19!} ]-1=\frac{21}{20!}-1 $
BETA
