- #1

Link

- 134

- 1

a = 2,6,12,20,30

S = 2+8+20+40...

(n^2) + n = (1/3) ((n^3)+3(n^2)+2n)

However, i cannot prove it by induction. It seems like there is smth wrong with the K^3.

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- Thread starter Link
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- #1

Link

- 134

- 1

a = 2,6,12,20,30

S = 2+8+20+40...

(n^2) + n = (1/3) ((n^3)+3(n^2)+2n)

However, i cannot prove it by induction. It seems like there is smth wrong with the K^3.

- #2

Muzza

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- #3

benorin

Homework Helper

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I think Link means [tex]\sum_{k=1}^{n} \left(k^2 + k\right) = \frac{1}{3}(n^3+3n^2+2n)[/tex], which is in fact true.Muzza said:

The inductive proof goes like this:

[tex]\sum_{k=1}^{n} \left(k^2 + k\right) = \frac{1}{3}(n^3+3n^2+2n)\Rightarrow \sum_{k=1}^{n} \left(k^2 + k\right) + \left((n+1)^2 + (n+1)\right) = \frac{1}{3}(n^3+3n^2+2n) + \left((n+1)^2 + (n+1)\right)[/tex]

[tex]\Rightarrow \sum_{k=1}^{n+1} \left(k^2 + k\right) = \frac{1}{3}(n^3+6n^2+11n+6) = \frac{1}{3}\left((n+1)^3+3(n+1)^2+2(n+1)\right) [/tex].

Therefore, [tex]\sum_{k=1}^{n} \left(k^2 + k\right) = \frac{1}{3}(n^3+3n^2+2n)[/tex] holds for every positive integer

Last edited:

- #4

Link

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Based on the results obtained above, how can i calculate 1^2 + 2^2 + 3^2...+n^2?

the problem is the word "calculate", do you think they want me to use the same method to derive a general expression for this series, or actually get an numerical value (i don't think its obtainable)? I am the only one in class doing this investigation and I cannot reach my teacher this week so advice on the meaning of this is appreciated.

- #5

HallsofIvy

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[tex]\sum_{k=1}^{n} \left(k^2 + k\right)=\sum_{k=1}^nk^2+ \sum_{k=1}^nk = \frac{1}{3}(n^3+3n^2+2n)[/tex]

If you know a formula for

[tex]\sum_{k=1}^{n} k[/tex]

just subtract that from both sides.

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