Let Sk, k = 1, 2, ......, 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{\left(100\right)^{2}}{100 !} + \sum_{k = 1}^{100} \left|\right. \textrm{ } \left(\right. k^{2} - 3 k + 1 \left.\right) S

Engineering

Mathematics

Special Types of Sequence

Question

Let S_k, k = 1, 2, ......, 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 $\frac{100^{2}}{100!} + \sum_{k = 1}^{100} | (k^{2} - 3 k + 1) S_{k} |$ , is

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$S_{k} = \frac{(k - 1)}{k!} \frac{k}{(k - 1)} = \frac{1}{(k - 1)!}$

$Now S = \frac{100^{2}}{100!} + \sum_{K = 1}^{100} \frac{| (k^{2} - 3 k + 1) |}{(k - 1)!} = \frac{100^{2}}{100!} + \sum_{K = 1}^{100} | \frac{(k - 1) (k - 2) - 1}{(k - 1)!} |$

$S = \frac{100^{2}}{100!} + \sum_{k = 3}^{100} | \frac{1}{(k - 3)!} - \frac{1}{(k - 1)!} |$

$Now T_{1} = \frac{1}{0!} = 1$

$T_{2} = \frac{1}{1!} = 1$

$T_{3} = \frac{1}{0!} - \frac{1}{2!}$

$T_{4} = \frac{1}{1!} - \frac{1}{3!}$

$T_{5} = \frac{1}{2!} - \frac{1}{4!}$

$T_{6} = \frac{1}{3!} - \frac{1}{5!}$

$⋮ ⋮ ⋮$

$T_{100} = \frac{1}{97!} - \frac{1}{99!}$

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$∴ S = \frac{100^{2}}{100!} + 4 - \frac{{(100)}^{2}}{100!} = 2 + 2 - (\frac{1}{98!} + \frac{1}{99!}) = 4 - (\frac{100}{99!}) = (4 - \frac{{(100)}^{2}}{100!}) = S = 4$

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