The sum of the series 4 1- - 11 2- - 22 3- - 37 4- - 56 5- - is

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Sum of n Terms

The sum of th...

Question

The sum of the series $\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + . . . .$ is

6 e

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6 e - 1

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5 e

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5 e + 1

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Solution

The correct option is C $6 e - 1$
Let $t_{n}$ be the $n$ th term of the series
$4 + 11 + 22 + 37 + 56 + . . .$
Let $S = 4 + 11 + 22 + . . . . + t_{n}$
$S = 4 + 11 + . . . . t_{n - 1} + t_{n}$
$\Rightarrow 0 = 4 + 7 + 11 + 15 + . . . . + t_{n - 1} + t_{n}$
$\Rightarrow t_{n} = 4 + [\frac{n - 1}{2} (2 \times 7 + (n - 2) 4))] = 2 n^{2} + n + 1$
Therefore, $S = \frac{\sum t_{n}}{n!} = \frac{2 \sum n^{2}}{n!} + \frac{\sum n}{n!} + \frac{\sum 1}{n!} = 2 (2 e) + e + (e - 1) = 6 e - 1$

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Similar questions

Q. Prove that

\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + . . . \infty = 6 e - 1

Q. Find the sum of the infinite series

\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + . . . .

\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + . . . \infty = b e - 1

Find b

$1 + \frac{1^{2} + 2^{2}}{2!} + \frac{1^{2} + 2^{2} + 3^{2}}{3!} + \frac{1^{2} + 2^{2} + 3^{2} + 4^{2}}{4!} + . . . \infty = \frac{b}{6} e$ . Find $b$

Q. The eccentricity of the conic 9x² + 25y² = 225 is
(a) 2/5
(b) 4/5
(c) 1/3
(d) 1/5
(e) 3/5

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The sum of the series 41!+112!+223!+374!+565!+.... is

The correct option is C 6e−1Let tn be the nth term of the series4+11+22+37+56+...Let S=4+11+22+....+tnS=4+11+....tn−1+tn⇒0=4+7+11+15+....+tn−1+tn⇒tn=4+[n−12(2×7+(n−2)4))]=2n2+n+1Therefore, S=∑tnn!=2∑n2n!+∑nn!+∑1n!=2(2e)+e+(e−1)=6e−1

The sum of the series $\frac{4}{1!} + \frac{11}{2!} + \frac{22}{3!} + \frac{37}{4!} + \frac{56}{5!} + . . . .$ is