Question

If \[1 + \left( {1 - {2^2} \times 1} \right) + \left( {1 - {4^2} \times 3} \right) + \left( {1 - {6^2} \times 5} \right) + ... + \left( {1 - {{20}^2} \times 19} \right) = \alpha - 220\beta \], then an ordered pair \[\left( {\alpha ,\beta } \right)\] is equal to
A. \[\left( {10,97} \right)\]
B. \[\left( {11,103} \right)\]
C. \[\left( {11,97} \right)\]
D. \[\left( {10,103} \right)\]

Answer 1

Hint: In this question, first by observing we calculate the \[{n^{th}}\] term of the series and then calculate the sum of the series by applying the formula of the sum of squares of natural numbers and the sum of cubes of natural numbers then compare LHS and RHS to find an ordered pair.

Formula used:
1. \[\sum {{n^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\]
2. \[\sum {{n^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\]

Complete step-by-step solution:
Given series is \[1 + \left( {1 - {2^2} \times 1} \right) + \left( {1 - {4^2} \times 3} \right) + \left( {1 - {6^2} \times 5} \right) + ... + \left( {1 - {{20}^2} \times 19} \right) = \alpha - 220\beta \]
Now we calculate the sum of series \[\left( {1 - {2^2} \times 1} \right) + \left( {1 - {4^2} \times 3} \right) + \left( {1 - {6^2} \times 5} \right) + ... + \left( {1 - {{20}^2} \times 19} \right)\]
Now we see that there are \[10\]terms in the above series.
By observation, we see that the \[{n^{th}}\] of a given series is
\[{T_n} = 1 - {\left( {2n} \right)^2}\left( {2n - 1} \right)\]
By simplifying, we get
\[
  {T_n} = 1 - 4{n^2}\left( {2n - 1} \right) \\
   = 1 - 8{n^3} + 4{n^2} \\
 \]
Therefore, the sum of the \[{n^{th}}\] term of the given series is \[{S_n} = \sum\limits_{n = 1}^{10} {{T_n}} \]
So, \[{S_n} = \sum\limits_{n = 1}^{10} {1 - 8{n^3} + 4{n^2}} \]
By splitting the terms, we get
\[
  {S_n} = \sum\limits_{n = 1}^{10} {1 - 8{n^3} + 4{n^2}} \\
   = \sum\limits_{n = 1}^{10} 1 + \sum\limits_{n = 1}^{10} {\left( { - 8{n^3}} \right)} + \sum\limits_{n = 1}^{10} {4{n^2}} \\
   = \sum\limits_{n = 1}^{10} 1 \,\, - 8\sum\limits_{n = 1}^{10} {{n^3}} + 4\sum\limits_{n = 1}^{10} {{n^2}} \,\,...\left( 1 \right) \\
 \]
We know that the sum of squares of first n natural numbers is given by \[\sum {{n^2}} = \dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}\] and the sum of cubes of first n natural numbers is given by \[\sum {{n^3}} = {\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2}\] \[{S_n} = n - 8{\left( {\dfrac{{n\left( {n + 1} \right)}}{2}} \right)^2} + 4\left( {\dfrac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right)\]
Now we know that \[n = 10\]
\[{S_{10}} = 10 - 8{\left( {\dfrac{{10\left( {10 + 1} \right)}}{2}} \right)^2} + 4\left( {\dfrac{{10\left( {10 + 1} \right)\left( {2.10 + 1} \right)}}{6}} \right)\]
Now by simplifying the above equation, we get
\[
  {S_{10}} = 10 - 8{\left( {\dfrac{{10\left( {10 + 1} \right)}}{2}} \right)^2} + 4\left( {\dfrac{{10\left( {10 + 1} \right)\left( {2.10 + 1} \right)}}{6}} \right) \\
   = 10 - 8{\left( {\dfrac{{10 \times 11}}{2}} \right)^2} + 4\left( {\dfrac{{10 \times 11 \times 21}}{6}} \right) \\
   = 10 - 8\dfrac{{{{\left( {10 \times 11} \right)}^2}}}{{{2^2}}} + \dfrac{4}{6}\left( {10 \times 11 \times 21} \right) \\
   = 10 - 8\dfrac{{{{\left( {110} \right)}^2}}}{4} + \dfrac{2}{3}\left( {110 \times 21} \right) \\
 \]
Further solving, we get
\[
  {S_{10}} = 10 - 2 \times 110 \times 110 + 2 \times 110 \times 7 \\
   = 10 - 2 \times 110\left( {110 - 7} \right) \\
   = 10 - 220\left( {103} \right) \\
 \]
Now the original given series is
\[1 + \left( {1 - {2^2} \times 1} \right) + \left( {1 - {4^2} \times 3} \right) + \left( {1 - {6^2} \times 5} \right) + ... + \left( {1 - {{20}^2} \times 19} \right) = \alpha - 220\beta \]
Now substitute the sum of series in it:
\[
  1 + 10 - 220\left( {103} \right) = \alpha - 220\beta \\
  11 - 220\left( {103} \right) = \alpha - 220\beta \\
 \]
By comparing LHS with RHS, we get
\[\alpha = 11,\,\,\,\beta = 103\]
Hence, option (B) is correct

Note: Students must use the summation formulas while calculating the sum because there are \[10\] terms then take the summation from \[1\] to \[10\]. Furthermore, the students should be able to compute the sum of squares and cubes of the first \[n\] natural numbers.