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Method of Differences (FP2)

Prove by the method of differences that

\displaystyle\sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1)

for n>1
Anyone can guide me on how to do this?

Reply 1

Kevin De Bruyne

Original post by JustDynamite

Prove by the method of differences that

\displaystyle\sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1)

for n>1
Anyone can guide me on how to do this?

Is this a real question? :s-smilie:

I can't see what the 'differences' would be.

Reply 2

B_9710

Maybe you just have to find someway of splitting it up such as

r(r+1)^2 - (r-1)r^2

and then finding the sum of the series

\sum^{n}_{r=1} \left( r(r+1)^2 -(r-1) r^2 \right )

and then rearranging it.

Reply 3

MathsSir

The method of differences is to look at the
sum to (k+1) - sum to k

1/6 (k+1)(k+2)(2(k+1)+1) - 1/6 k(k+1)(2K+1)
= 1/6 (k+1) [(k+2)(2k+3) - k(2k+1)]
= 1/6 (k+1)[(2k^2+7k+6 -2k^2 -k]
= 1/6 (k+1) [6k+6]
= (k+1)^2

hope this is reasonably clear.

Reply 4

Zacken

Original post by JustDynamite

Prove by the method of differences that

\displaystyle\sum_{r=1}^n r^2 = \frac{1}{6}n(n+1)(2n+1)

for n>1
Anyone can guide me on how to do this?

Original post by SeanFM

Is this a real question? :s-smilie:

I can't see what the 'differences' would be.

That's correct Sean.

Below is the full question:

Reply 5

Kevin De Bruyne

Original post by Zacken

That's correct Sean.

Below is the full question:

well remembered.

@JustDynamite, look at the link between question.

Reply 6

Alexion

Original post by MathsSir

... that's not the method of differences, pretty sure that's proof by induction :/

Reply 7

JustDynamite

Ah thanks guys! It was question 24 from here: https://8dedc505ac3fba908c50836f59059ccce5cd0f1e.googledrive.com/host/0B1ZiqBksUHNYdHIxUkJmdndfMlE/Questions%20from%20Old%20Papers%20-%20FP2%20Edexcel.pdf

Looks like the top part was left out, was able to do it using the part Zacken shown

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Method of Differences (FP2)

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