Sum problem

Question

0

323

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The sum 1^2 + 2^2 + 3^2 + ... + n^2 = n(n + 1)(2n + 1)/6.

What is the value of 21^2 + 22^2 + 23^2 + ... + 40^2 + 41^2 + 42^2?

Guest Dec 18, 2021

Guest · Answer 1 · 2021-12-19T12:27:24

What is the value of 21^2 + 22^2 + 23^2 + ... + 40^2 + 41^2 + 42^2==22,715

geno3141 · Answer 2 · 2021-12-19T01:05:29

Using the formulas: 1 + 2 + ... + n = (n(n + 1)/2 and 1² + 2² + ... + n² = n(n + 1)(2n + 1)/6

to find the value of: 21² + 22² + ... + 41² + 42²:

Note that there are 22 addends.

21² + 22² + ... + 41² + 42² =

(20 + 1)² + (20 + 2)² + (20 + 21)² + (20 + 22)²

Using the binomial expansion:

= ( 20² + 2·20·1 + 1² ) + ( 20² + 2·20·2 + 2² ) + ... + ( 20² + 2·20·21 + 21² ) + ( 20² + 2·20·22 + 22² )

Rearranging:

= [ 20² + 20² + ... + 20² + 20² ] + [ 2·20·1 + 2·20·2 + ... + 2·20·21 + 2·20·22 ] + [ 1² + 2² + ... + 21² + 22² ]

= 22[ 20² ] + [ 2·20·1 + 2·20·2 + ... + 2·20·21 + 2·20·22 ] + [ 1² + 2² + ... + 21² + 22² ]

= 8800 + [ 2·20·1 + 2·20·2 + ... + 2·20·21 + 2·20·22 ] + [ 1² + 2² + ... + 21² + 22² ]

= 8800 + 40 · [ 1 + 2 + ... + 21 + 22 ] + [ 1² + 2² + ... + 21² + 22² ]

= 8800 + 40 · [ 22(1 + 22)/2 ] + [ 1² + 2² + ... + 21² + 22² ]

= 8800 + 40 · [ 253 ] + [ 1² + 2² + ... + 21² + 22² ]

= 8800 + 10120 + [ 1² + 2² + ... + 21² + 22² ]

= 8800 + 10120 + [ 22(22 + 1)(2·22 + 1)/6 ]

= 8800 + 10120 + 3795

= 22715