Stuck on proving series question

This question seems quite strange. Chapter in book is FP1 induction and series.

Using the idea suggested in the diagram prove that

1n + 2(n - 1) + 3(n - 2) + ... + n times 1 = \frac{1}{6}n(n+1)(n+2)


1   1   1   1   1                      1   |   1   |   1   |   1   |   1
-------------------------                      |       |        |        |
                                               |       |        |        |
2   2    2   2                         2    |   2  |    2   |   2   |
--------------------                            |       |        |        |
                                                |       |        |
3   3    3                  =          3    |   3  |    3  |
----------------                                |       |        |
                                                |       |        |
4    4                                   4    |   4  |
----------                                      |       |
                                                |       |
5                                         5    |

Thinking about the series I came up with:

\sum\limits_{r=1}^{n} r(n - r + 1)

\sum\limits_{r=1}^{n} rn - \sum\limits_{r=1}^{n} r^2 + \sum\limits_{r=1}^{n} r

= n^2 \sum\limits_{r=1}^{n} r - \sum\limits_{r=1}^{n} r^2 + \sum\limits_{r=1}^{n} r

I am confused about the diagram and how that is supposed to help???

Reply 1

SherlockHolmes

13

Original post by acomber


1   1   1   1   1                      1   |   1   |   1   |   1   |   1
-------------------------                      |       |        |        |
                                               |       |        |        |
2   2    2   2                         2    |   2  |    2   |   2   |
--------------------                            |       |        |        |
                                                |       |        |
3   3    3                  =          3    |   3  |    3  |
----------------                                |       |        |
                                                |       |        |
4    4                                   4    |   4  |
----------                                      |       |
                                                |       |
5                                         5    |

If you let n=5 then you would get the diagram above.

From the first term in the series you will get five 1s, second term will give you four 2s etc.

Reply 2

acomber

OP

11

Sorry, I am not getting that?

If n = 5 then

\sum\limits_{r=1}^{5} r(5 - r + 1)

First term:

1(5 - 1 + 1) = 5

Original post by SherlockHolmes

If you let n=5 then you would get the diagram above.

From the first term in the series you will get five 1s, second term will give you four 2s etc.

Reply 3

SherlockHolmes

13

Original post by acomber

Sorry, I am not getting that?

If n = 5 then

\sum\limits_{r=1}^{5} r(5 - r + 1)

First term:

1(5 - 1 + 1) = 5

Yes, and that is equal to the sum of the numbers in the first row of your diagram.

I don't think the diagram is too important to be honest because you have already worked out what the summation should be (which is correct).

However, I think your last line of working is incorrect. You should take out a factor of n from the first summation but you have taken out a factor of n^2. So it should be:

n \sum\limits_{r=1}^{n} r - \sum\limits_{r=1}^{n} r^2 + \sum\limits_{r=1}^{n} r

Reply 4

acomber

OP

11

Ah ok, thanks for that help.

My thinking on the sum was that

\sum\limits_{r=1}^{n} rn

is the sum of r x n the sum of r is obviously sum ( r ) and sum n == n x n because n is constant to be summed n times, (n + n + ... + n n times) so altogether I get:

= n^2 \sum\limits_{r=1}^{n} r

No?

Ah, But it is the factor multiplied by sum ( r ) , yes so n sum ( r ) == sum ( rn ) . I think I get it now.

So using

n \sum\limits_{r=1}^{n} r - \sum\limits_{r=1}^{n} r^2 + \sum\limits_{r=1}^{n} r

= n(\frac{1}{2}n(n+1)) - \frac{1}{6}n(n+1)(2n+1) + \frac{1}{2}n(n+1)

= \frac{1}{6}n(n+1)[ 3n - (2n+1) + 3]

= \frac{1}{6}n(n+1)(n+2) = RHS

Perfect. Thank you.

Original post by SherlockHolmes

Yes, and that is equal to the sum of the numbers in the first row of your diagram.

I don't think the diagram is too important to be honest because you have already worked out what the summation should be (which is correct).

However, I think your last line of working is incorrect. You should take out a factor of n from the first summation but you have taken out a factor of n^2. So it should be: