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# Method of Differences (FP2)

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1. Prove by the method of differences that
for n>1
Anyone can guide me on how to do this?
2. (Original post by JustDynamite)
Prove by the method of differences that
for n>1
Anyone can guide me on how to do this?
Is this a real question? I can't see what the 'differences' would be.
3. Maybe you just have to find someway of splitting it up such as and then finding the sum of the series and then rearranging it.
4. 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.
5. (Original post by JustDynamite)
Prove by the method of differences that
for n>1
Anyone can guide me on how to do this?
(Original post by SeanFM)
Is this a real question? I can't see what the 'differences' would be.
That's correct Sean.

Below is the full question:

6. (Original post by Zacken)
That's correct Sean.

Below is the full question:

well remembered.

JustDynamite, look at the link between question.
7. (Original post by 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.
... that's not the method of differences, pretty sure that's proof by induction :/
8. Ah thanks guys! It was question 24 from here: https://8dedc505ac3fba908c50836f5905...%20Edexcel.pdf

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

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