Question

Let we are given a expression of summation ${S_n} = \sum\limits_{k = 1}^{4n} {{{\left( { - 1} \right)}^{\dfrac{{k\left( {k + 1} \right)}}{2}}}{k^2}} $ then ${S_n}$ can take value(s)
$\left( a \right)1056$
$\left( b \right)1088$
$\left( c \right)1120$
$\left( d \right)1332$

Answer 1

Hint: In this particular question use the concept that first expands the summation and arranges them such that it becomes the sum of two arithmetic progression, so use these concepts to reach the solution of the question.

Complete step-by-step solution:
Given equation
${S_n} = \sum\limits_{k = 1}^{4n} {{{\left( { - 1} \right)}^{\dfrac{{k\left( {k + 1} \right)}}{2}}}{k^2}} $
Now expand this summation we have,
$ \Rightarrow {S_n} = {\left( { - 1} \right)^{\dfrac{{1\left( {1 + 1} \right)}}{2}}}{1^2} + {\left( { - 1} \right)^{\dfrac{{2\left( {2 + 1} \right)}}{2}}}{2^2} + {\left( { - 1} \right)^{\dfrac{{3\left( {3 + 1} \right)}}{2}}}{3^2} + {\left( { - 1} \right)^{\dfrac{{4\left( {4 + 1} \right)}}{2}}}{4^2} + {\left( { - 1} \right)^{\dfrac{{5\left( {5 + 1} \right)}}{2}}}{5^2}...... + {\left( { - 1} \right)^{\dfrac{{4n\left( {4n + 1} \right)}}{2}}}{\left( {4n} \right)^2}$
$ \Rightarrow {S_n} = - {1^2} - {2^2} + {3^2} + {4^2} - {5^2} - ...... + {\left( {4n} \right)^2}$, total 4n terms
Above series is also written as,
$ \Rightarrow {S_n} = \left( {{3^2} - {1^2}} \right) + \left( {{4^2} - {2^2}} \right) + \left( {{7^2} - {5^2}} \right) + \left( {{8^2} - {6^2}} \right) + ......$ Up to 2n terms
$ \Rightarrow {S_n} = \left( 8 \right) + \left( {12} \right) + \left( {24} \right) + \left( {28} \right) + ......$ Up to 2n terms
$ \Rightarrow {S_n} = 2\left[ {\left( 4 \right) + \left( 6 \right) + \left( {12} \right) + \left( {14} \right) + ......{\text{up to 2n terms}}} \right]$
$ \Rightarrow {S_n} = 2\left\{ {\left( {4 + 12 + 20 + .......{\text{ up to n terms}}} \right) + \left( {6 + 14 + 22 + .....{\text{up to n terms}}} \right)} \right\}$
Now as we see that both of the above series forms an A.P with first term 4 and 6 respectively, common difference, (12 – 4) = (20 – 12) = 8, and (14 – 6) = (22 – 14) = 8 respectively, and the number of terms in both if the above series are same i.e. n terms.
Now as we know that the sum of the A.P is given as,
$S = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where symbols have their usual meanings.
$ \Rightarrow {S_n} = 2\left\{ {\left( {\dfrac{n}{2}\left( {2\left( 4 \right) + \left( {n - 1} \right)8} \right)} \right) + \left( {\dfrac{n}{2}\left( {2\left( 6 \right) + \left( {n - 1} \right)8} \right)} \right)} \right\}$
Now simplify this we have,
$ \Rightarrow {S_n} = 2\left\{ {\left( {n\left( {4 + \left( {n - 1} \right)4} \right)} \right) + \left( {n\left( {6 + \left( {n - 1} \right)4} \right)} \right)} \right\}$
$ \Rightarrow {S_n} = 2\left\{ {\left( {n\left( {4n} \right)} \right) + \left( {n\left( {4n + 2} \right)} \right)} \right\}$
$ \Rightarrow {S_n} = 2\left\{ {4{n^2} + 4{n^2} + 2n} \right\}$
$ \Rightarrow {S_n} = 2\left\{ {8{n^2} + 2n} \right\}$
$ \Rightarrow {S_n} = 16{n^2} + 4n$............ (1)
Now we have to find out the values which ${S_n}$ can take from the given options so, substitute, n = 1, 2, 3, 4..... and so on we have,
Now substitute n = 1 in equation (1) we have,
$ \Rightarrow {S_1} = 16{\left( 1 \right)^2} + 4\left( 1 \right) = 20$
Now as we see that given options has a higher value so it is possible for higher value of n,
So substitute n = 7 in equation (1) we have,
$ \Rightarrow {S_7} = 16{\left( 7 \right)^2} + 4\left( 7 \right) = 812$
So this also not the values which is given in the options so again substitute higher value of n in equation (1) we have,
So, substitute n = 8 in equation (1) we have,
$ \Rightarrow {S_8} = 16{\left( 8 \right)^2} + 4\left( 8 \right) = 1056$
Now substitute n = 9 in equation (1) we have,
$ \Rightarrow {S_9} = 16{\left( 9 \right)^2} + 4\left( 9 \right) = 1332$
So this is the highest value given in the options.
Hence options (a) and (d) both are the correct answers, which ${S_n}$ can take.
Hence options (a) and (d) both are the correct answers.

Note: Whenever we face such types of questions the key concept we have to remember is that always recall the formula of sum of n terms of an A.P which is given as ${S_n} = \dfrac{n}{2}\left( {2a + \left( {n - 1} \right)d} \right)$, where n is the number of terms, a is the first term and d is the common difference of the series.