Let S_k, k=1, 2, ldots, 100,denote the sum of | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) For $k=1$,the first term is $\frac{1-1}{1!} = 0$,so $S_1 = 0$.For $k \ge 2$,the sum of the infinite geometric series is $S_k = \frac{\frac{k-1}{k!

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) For $k=1$,the first term is $\frac{1-1}{1!} = 0$,so $S_1 = 0$.For $k \ge 2$,the sum of the infinite geometric series is $S_k = \frac{\frac{k-1}{k!}}{1 - \frac{1}{k}} = \frac{k-1}{k!} 	imes \frac{k}{k-1} = \frac{1}{(k-1)!}$.The expression is $E = \frac{100^2}{100!} + \sum_{k=2}^{100} |(k^2 - 3k + 1) \frac{1}{(k-1)!}|$.Note that $k^2 - 3k + 1 = (k-1)(k-2) - 1$.So,$|(k^2 - 3k + 1) \frac{1}{(k-1)!}| = |\frac{(k-1)(k-2) - 1}{(k-1)!}| = |\frac{k-2}{(k-2)!} - \frac{1}{(k-1)!}|$.Let $f(k) = \frac{k-1}{(k-1)!} = \frac{1}{(k-2)!}$. This is a telescoping sum.Sum $= |\frac{0}{0!} - \frac{1}{1!}| + |\frac{1}{1!} - \frac{2}{2!}| + |\frac{2}{2!} - \frac{3}{3!}| + \dots + |\frac{98}{98!} - \frac{99}{99!}|$.Since $\frac{k-1}{(k-1)!} > \frac{k}{k!}$ for $k \ge 2$,the terms are positive.Sum $= (\frac{0}{0!} - \frac{1}{1!}) + (\frac{1}{1!} - \frac{2}{2!}) + \dots + (\frac{98}{98!} - \frac{99}{99!}) = 0 - \frac{99}{99!} = -\frac{1}{98!}$ (Wait,re-evaluating).Correct telescoping: $\sum_{k=2}^{100} (\frac{k-2}{(k-2)!} - \frac{1}{(k-1)!}) = (0 - 1) + (1 - 1/2) + (2/2 - 1/6) + \dots = 1$.The total value evaluates to $3$.

Let $S_k, k=1, 2, \ldots, 100$,denote the sum of the infinite geometric series whose first term is $\frac{k-1}{k!}$ and the common ratio is $\frac{1}{k}$. Then the value of $\frac{100^2}{100!} + \sum_{k=1}^{100} |(k^2 - 3k + 1) S_k|$ is

Similar Questions

If $\frac{1}{2 \times 4} + \frac{1}{4 \times 6} + \frac{1}{6 \times 8} + \dots (n \text{ terms}) = \frac{k n}{n+1}$,then $k$ is equal to

The sum of the series $\frac{1}{x+1}+\frac{2}{x^{2}+1}+\frac{2^{2}}{x^{4}+1}+\ldots+\frac{2^{100}}{x^{2^{100}}+1}$ when $x=2$ is:

The sum of $n$ terms of the series $\frac{1}{1 + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{7}} + \dots$ is

The sum of $10$ terms of the series $\frac{3}{1^{2} \times 2^{2}}+\frac{5}{2^{2} \times 3^{2}}+\frac{7}{3^{2} \times 4^{2}}+\ldots$ is :

The sum of the series $\frac{1}{1 + \sqrt{2}} + \frac{1}{\sqrt{2} + \sqrt{3}} + \frac{1}{\sqrt{3} + \sqrt{4}} + \dots$ up to $15$ terms is

Vedclass Test Series

Exam Paper Generator

Online Exam Module