Question

Let $S$ denote the sum of the infinite series $1+\frac{8}{2!}+\frac{21}{3!}+\frac{40}{4!}+\frac{65}{5!}+.......$. Then

• A. $S<8$
• B. $S>12$
• C. $8<S<12$
• D. $S=8$

Answer 1

Answer: (C)

$8<S<12$

$S_n=1+8+21+40+65+.....+a_n-\underset{–}{S_n=1+8+21+40+.....+a_{n-1}+a_n}$

$0=1+7+13+19+25+...,,,,,,,,.-a_n$

$a_n=1+7+13+19+25+....a_n=\frac{n}{2}[2a+(n-1\left)d\right]=\frac{n}{2}[2+(n-1\left)6\right]a_n=n(3n-2)S={\sum }_{i=1}^{\infty }\frac{n(3n-2)}{n!}={\sum }_{i=1}^{\infty }\frac{(3n-2)}{(n-1)!}$

Adding and subtracting $1$ in numerator

$S={\sum }_{i=2}^{\infty }\frac{3(n-1)}{(n-1)!}+{\sum }_{i=1}^{\infty }\frac{1}{(n-1)!}S=3{\sum }_{i=2}^{\infty }\frac{1}{(n-2)!}+{\sum }_{i=1}^{\infty }\frac{1}{(n-1)!}S=3e+eS=4e8<S<12e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+....\infty e=1+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+....\infty$