Question

Let \[{{S}_{k}},k=1,2,...,100\], denote the sum of the infinite geometric series whose first term is \[\dfrac{k-1}{k!}\] and the common ratio is \[\dfrac{1}{k}\].
Then, the value of \[\dfrac{{{100}^{2}}}{100!}+\sum\limits_{k=2}^{100}{\left| \left( {{k}^{2}}-3k+1 \right){{S}_{k}} \right|}\] is:
(). 3
(). 1
(). 2
(). 4

Answer 1

Hint: Use the formula of sum of infinite geometric series to find the \[{{S}_{k}}\]. Substitute the \[{{S}_{k}}\] in given expression and then simplify it algebraically like normal algebraic equation, the sum is given by:
\[S=\dfrac{a}{1-r}\], where a is first term and r is common ratio of the infinite geometric series, for r < 1 this is true. So in the given question you have the value of first term and common ratio. Find the value of the sum of infinite terms by substituting these into the formula. Then use normal algebra to solve the sum expression you get. Try to convert the expression you have into the form of expression in the question.

Complete step-by-step answer:

Given expression in the question for which we need the value:
\[\dfrac{{{100}^{2}}}{100!}+\sum\limits_{k=2}^{100}{\left| \left( {{k}^{2}}-3k+1 \right){{S}_{k}} \right|}\]
Take the term in the summation separately and solve it,
\[\sum\limits_{k=2}^{100}{\left| \left( {{k}^{2}}-3k+1 \right){{S}_{k}} \right|}\]
For solving the above term, we need the value of term \[{{S}_{k}}\]. By the basic knowledge of sequences we know the formula for sum of terms in infinite series with first term a, common ratio r, of an infinite geometric series with sum S is,
\[S=\dfrac{a}{1-r}\]
Here, \[S={{S}_{k}},a=\dfrac{k-1}{k!},r=\dfrac{1}{k}\].
By substituting values of S, a, r into sum formula, we get:
\[{{S}_{k}}=\dfrac{\dfrac{k-1}{k!}}{1-\dfrac{1}{k}}=\dfrac{1}{\left( k-1 \right)!}\], for k > 1
By substituting above equations in the summation term, we get:
\[\sum\limits_{k=2}^{100}{\left| \left( {{k}^{2}}-3k+1 \right)\dfrac{1}{\left( k-1 \right)!} \right|}\]
By basic knowledge of algebra, we know that this is true:
\[{{k}^{2}}-3k+1={{\left( k-1 \right)}^{2}}-k\]
By substituting this into the summation term, we get:
\[=\sum\limits_{k=2}^{100}{\left| \dfrac{{{\left( k-1 \right)}^{2}}-k}{\left( k-1 \right)!} \right|}\]
By turning this dfraction in the modulus in 2 dfractions, we get:
\[=\sum\limits_{k=2}^{100}{\left| \dfrac{\left( k-1 \right)}{\left( k-2 \right)!}-\dfrac{k}{\left( k-1 \right)!} \right|}\]
By substituting values of k one by one into the terms, we get:
\[=\left| \dfrac{1}{0!}-\dfrac{2}{1!} \right|+\left| \dfrac{2}{1!}-\dfrac{3}{2!} \right|+\left| \dfrac{3}{2!}-\dfrac{4}{3!} \right|+......\]
By substituting the modulus in each term above, we get:
\[=\dfrac{2}{1!}-\dfrac{1}{0!}+\dfrac{2}{1!}-\dfrac{3}{2!}-\dfrac{4}{3!}+......+\dfrac{99}{98!}-\dfrac{100}{99!}\]
By telescopic cancellation of all terms in above equation, we get:
\[=-1+2.\dfrac{2}{1!}-\dfrac{100}{99!}\]
By simplifying more and then equating it to summation term, we get
\[=3-\dfrac{100}{99!}\]
Multiply and divide by 100 in the term with minus sign, we get:
\[\sum\limits_{k=2}^{100}{\left| \left( {{k}^{2}}-3k+1 \right){{S}_{k}} \right|}=3-\dfrac{{{100}^{2}}}{100!}\]
By adding the term the minus sign on both sides, we get:
\[\dfrac{{{100}^{2}}}{100!}+\sum\limits_{k=2}^{100}{\left| \left( {{k}^{2}}-3k+1 \right){{S}_{k}} \right|}=3-\dfrac{{{100}^{2}}}{100!}+\dfrac{{{100}^{2}}}{100!}\]
By cancelling the term which is in common, we conclude to get 3.
\[\dfrac{{{100}^{2}}}{100!}+\sum\limits_{k=2}^{100}{\left| \left( {{k}^{2}}-3k+1 \right){{S}_{k}} \right|}=3\]
3 is the value of required expression.
So, option (a) is correct.

Note: Be careful while calculating the summation term if you mess it up with changing one sign the whole answer might change. Be careful while taking out the modulus, you need to observe that the first term is negative so you need to write negative expressions inside the modulus before removing the modulus. Whenever you have infinite sum the main idea you need to get is telescopic cancellation. Because it is the only way which converts any infinite sum into a sum of one or two terms.