
Find the sum of the series \[{11^2} + {12^2} + {13^2} + ... + {20^2}\] .
A.2481
B.2483
C.2485
D.2487
Answer
163.8k+ views
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.
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.
Recently Updated Pages
Geometry of Complex Numbers – Topics, Reception, Audience and Related Readings

JEE Main 2021 July 25 Shift 1 Question Paper with Answer Key

JEE Main 2021 July 22 Shift 2 Question Paper with Answer Key

JEE Atomic Structure and Chemical Bonding important Concepts and Tips

JEE Amino Acids and Peptides Important Concepts and Tips for Exam Preparation

JEE Electricity and Magnetism Important Concepts and Tips for Exam Preparation

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

Atomic Structure - Electrons, Protons, Neutrons and Atomic Models

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Displacement-Time Graph and Velocity-Time Graph for JEE

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Degree of Dissociation and Its Formula With Solved Example for JEE

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

Instantaneous Velocity - Formula based Examples for JEE

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations

NCERT Solutions for Class 11 Maths Chapter 8 Sequences and Series
