Question

Find the sum of the series \[{11^2} + {12^2} + {13^2} + ... + {20^2}\] .
A.2481
B.2483
C.2485
D.2487

Answer 1

Hint: Observe that in the given series there are 10 terms but our formula starts from 1 so first obtain the sum of square of first 20 terms then subtract the sum of first ten term to obtain the result.

Formula Used: The formula of sum of n square terms \[{1^2} + {2^2} + {3^2} + ... + {n^2}\] is \[\dfrac{{n(n + 1)(2n + 1)}}{6}\] .

Complete step by step solution: Observe that in the given series there are 10 terms but our formula starts from 1 so first obtain the sum of square of first 20 terms then subtract the sum of first ten term to obtain the result.
Substitute 20 for n in the formula \[\dfrac{{n(n + 1)(2n + 1)}}{6}\] and calculate to obtain the required result.
\[\dfrac{{20(20 + 1)(2.20 + 1)}}{6}\]
\[ = \dfrac{{20 \times 21 \times 41}}{6}\]
=2870
Substitute 10 for n in the formula \[\dfrac{{n(n + 1)(2n + 1)}}{6}\] and calculate to obtain the required result.
\[\dfrac{{10(10 + 1)(2.10 + 1)}}{6}\]
\[ = \dfrac{{10 \times 11 \times 21}}{6}\]
=385
Subtract 385 from 2870 to obtain the required result.
\[2870 - 385 = 2485\]

 Option ‘C’ is correct

Note: Sometime students directly substitute 10 for n and calculate the result but observe that our formula start from 1 so we need to calculate the sum of square of first 20 terms first then have to subtract the sum of square of first 10 terms to obtain the required result.