A number theory problem by A Former Brilliant Member

Number Theory Level 2

The value of 1 1 2 + 1 2 2 + 1 3 2 + … … … … + 2 0 2 11^{2}+12^{2}+13^{2}+…………+20^{2} is?

2870 2485 385 3255

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A Former Brilliant Member by A Former Brilliant Member

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2 solutions

11^2+12^+13^2+……+20^2.
=[1^2+2^2+3^2+……+20^2]- [1^2+2^2+3^2+……+10^2] =[20(20+1)(40+1)- 10(10+1)(20+1)]/6 =2485

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The formula of sum of squares from 1 to n is n ( n + 1 ) ( 2 n + 1 ) 6 . . . . . . We have (1 to 20) - (1 to 10). ∴ r e q u i r e d v a l u e i s = 20 ∗ 21 ∗ 41 − 10 ∗ 11 ∗ 21 6 = 210 ( 2 ∗ 41 − 11 ) 6 = 35 ∗ 71 = 2485. \text{The formula of sum of squares from 1 to n is}\\ \dfrac{n(n+1)(2n+1)}{6}......\text{We have (1 to 20) - (1 to 10).}\\\therefore~required~value~ is\\=\dfrac{20*21*41 - 10*11*21}{6}= \dfrac{210(2*41 - 11)}{6}= 35*71=2485.

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