1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2-3n-1)/2

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04-07-2016
Mathematics

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1^2 + 4^2 + 7^2 + ... + (3n - 2)^2 = n(6n^2-3n-1)/2

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jhonyy9 jhonyy9 · Answer 1 · 2016-07-04T22:46:11+02:00

jhonyy9 jhonyy9

04-07-2016

use math induction proof method will be usefully

for n=1 will get

1^2 = 1(6*1^2 -3*1 -1)/2
1 = 1(6-4)/2
1 = 1(2)/2
1 = 1

suppose true for n=k and prove for k = k+1

hope this will help you

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