Question

The sum of the series \[4 + 16 + 36 + 64 + ........ + 400\] is
A. \[1530\]
B. \[1540\]
C. \[1550\]
D. \[1560\]

Answer 1

Hint: We take the value 4 from each number of the series. Then we see that the series in the bracket is a series of squares of natural numbers. Using the formula for the sum of squares of n terms we calculate the sum and multiply it by 4.
* Sum of series \[{1^2} + {2^2} + {3^2} + ......... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}\].

Complete step-by-step answer:
We are given the series \[4 + 16 + 36 + 64 + ........ + 400\].
We can write the terms of the series as
\[4 = 4 \times 1\]
\[16 = 4 \times 4\]
\[36 = 4 \times 9\]
.
.
.
\[400 = 4 \times 100\]
Since we see that all numbers of the series are factors of 4, we can take 4 common and write the terms of series in the bracket.
\[ \Rightarrow 4 \times \{ 1 + 4 + 9 + 16 + ........ + 100\} \]
Now we can write the terms in the bracket as \[1 = {1^2};4 = {2^2};9 = {3^2},16 = {4^2}.........100 = {10^2}\].
\[ \Rightarrow 4 \times \{ {1^2} + {2^2} + {3^2} + {4^2} + ......... + {10^2}\} \] … (1)
We see that the bracket contains a sum of squares of the first ten numbers.
So, we apply the formula for the sum of squares of first n natural numbers.
\[{1^2} + {2^2} + {3^2} + ......... + {n^2} = \dfrac{{n(n + 1)(2n + 1)}}{6}\]
Since the number of terms is 10. Substitute the value of n=10 in the formula.
\[
   \Rightarrow {1^2} + {2^2} + {3^2} + ......... + {10^2} = \dfrac{{10(10 + 1)(2(10) + 1)}}{6} \\
   \Rightarrow {1^2} + {2^2} + {3^2} + ......... + {10^2} = \dfrac{{10 \times (11) \times (21)}}{{2 \times 3}} \\
 \]
Cancel out the same terms from numerator and denominator.
\[
   \Rightarrow {1^2} + {2^2} + {3^2} + ......... + {10^2} = \dfrac{{(2 \times 5) \times (11) \times (3 \times 7)}}{{2 \times 3}} \\
   \Rightarrow {1^2} + {2^2} + {3^2} + ......... + {10^2} = (5) \times (11) \times (7) \\
   \Rightarrow {1^2} + {2^2} + {3^2} + ......... + {10^2} = 385 \\
 \]
Now we substitute the value of sum in equation (1).
\[ \Rightarrow 4 \times \{ {1^2} + {2^2} + {3^2} + {4^2} + ......... + {10^2}\} = 4 \times 385 = 1540\]
So, the sum of series \[4 + 16 + 36 + 64 + ........ + 400\] is \[1540\]

So, the correct answer is “Option B”.

Additional Information:
* Sum of first n natural numbers is given by \[\{ 1 + 2 + 3 + ........ + n\} = \dfrac{{n(n + 1)}}{2}\]
* Sum of cubes of first n natural numbers is given by \[{1^3} + {2^3} + {3^3} + ........ + {n^3} = {\left( {\dfrac{{n(n + 1)}}{2}} \right)^2}\]

Note: Students might make the mistake of solving the value inside the bracket by normal addition which is a very complicated and long process. Students are very likely to make calculation mistakes when adding so many values, so try to avoid calculation by adding these many terms.