jenks88 wrote:
I get the general term equation but still unsure where ∑\(n^{2}\)=\(\frac{n(n+1)(2n+1)}{6}\) comes from?
Hi Jenks88,
I would suggest to memorize the below formulae :-
Arithmetic progression:-
1. 1 + 2 + 3 + ...... + n = [n(n + 1)]/2.
2.1² + 2² + 3² +……………. + n² = [n(n+ 1)(2n+ 1)]/6.
3.1³ + 2³ + 3³ + . . . . + n³ = [{n(n + 1)}/2 ]
Geometric Progression:-
(i) The general form of a G.P. is a, ar, ar², ar³, . . . . . where a is the first term and r, the common ratio of the G.P.
(ii) The n th term of the above G.P. is t₀ = a.rn−1 .
(iii) The sum of first n terms of the above G.P. is S = a ∙ [(1 - rⁿ)/(1 – r)] when -1 < r < 1
or, S = a ∙ [(rⁿ – 1)/(r – 1) ]when r > 1 or r < -1.
(iv) The geometric mean of two positive numbers a and b is √(ab) or, -√(ab).
(v) a + ar + ar² + ……………. ∞ = a/(1 – r) where (-1 < r < 1).
Also, practice several sums on summation & series so that you would be able to master the application of the above formulae.
Hope it helps.
_________________
Regards,
PKN
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50% (02:10) correct 
