Answers only work once. In exams it is methods that count.
To find the sum of the series 2,6,12,20... given nth term is n(n+1)
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Method 1: Experimentation
Compare partial sums with individual terms
2, 8, 20, 40, 70, 112
2, 6, 12, 20, 30, 42
Since denominator is often 3 express all these in
terms of thirds and observe how ratio depends on n
3/3, 4/3, 5/3, 6/3, 7/3, 8/3
3 is 1 +2,
4 is 2 + 2
5 is 3 + 2 and so on.
Sum to n terms, S = [(n + 2)/3] * nth term
Sum to n terms, S = n(n+1)(n + 2)/3
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Although that was easy to understand, it relied on the (unlikely?) chance of finding a simple relation between partial sums and nth term. A more reliable approach uses the SIGMA notation.
Method 2: The ith term is i^2 + i so what we want is the sum S, where
S = (i = 1 to n)SIGMA [i^2 + i]. The 1 to n part is understood in the next few SIGMAS
SIGMA[(i + 1)^3 - i^3] = SIGMA(3i^2 + 3i) + SIGMA (1) = 3S + n ...(a)
The justification of introducing that quantity is that it "telescopes"
SIGMA[(i + 1)^3 - i^3] = (2^3 - 1^3) + (3^3 - 2^3) + (4^3 - 3^3) + ... + [(n + 1)^3 - n^3]
so that all terms except the first and last cancel out
SIGMA[(i + 1)^3 - i^3] = (n + 1)^3 - 1 ........................(b)
Comparing a) and b)
3S = n^3 + 3n^2 + 3n + 1 - 1 - n = n^3 + 3n^2 + 2n which rearranges as
Sum to n terms, S = n(n+1)(n + 2)/3
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Regards - Ian (YA)
