FP1 Summation Series when r=0

I don't know what to do when you get a summation series where r=0 :frown:

To be more specific, question 10ii, paper June 2013 (R) from Edexcel:
https://gyazo.com/501ce98024b3cbff43cdc682dc6bb5a2
(above is the screenshot of the question)

I've read through the mark scheme but I genuinely don't understand
https://gyazo.com/2425b703b74bbd62fe5cb2fb2e8d3c45

Could anyone please explain what you have to do when r=0, and what do you do when you are given a series with 'n' in (that is 'n' mixed up with 'r'). I know how to expand with 'r's, but with 'n'... I don't know :frown:

Thank you!

Reply 1

lerjj

13

Original post by Horagontus

I don't know what to do when you get a summation series where r=0 :frown:

To be more specific, question 10ii, paper June 2013 (R) from Edexcel:
https://gyazo.com/501ce98024b3cbff43cdc682dc6bb5a2
(above is the screenshot of the question)

I've read through the mark scheme but I genuinely don't understand
https://gyazo.com/2425b703b74bbd62fe5cb2fb2e8d3c45

Could anyone please explain what you have to do when r=0, and what do you do when you are given a series with 'n' in (that is 'n' mixed up with 'r' :wink:

. I know how to expand with 'r's, but with 'n'... I don't know :frown:

Thank you!

First split it up into three seperate sums:

\displaystyle \sum^n_{r=0} r^2-2 \sum^n_{r=0} r + \sum^n_{r=0} (2n+1)

Note that the first two sums have the leading term = 0 because 0^2=0 and 0=0. The last term is simply the sum of (n+1) terms all of which each the constant value (2n+1).

Reply 2

Zacken

22

Original post by Horagontus

Could anyone please explain what you have to do when r=0, and what do you do when you are given a series with 'n' in (that is 'n' mixed up with 'r' :wink:

. I know how to expand with 'r's, but with 'n'... I don't know :frown:

Thank you!

Two ways to look at it:

1.

\displaystyle \sum_{r=0}^n (2n+1) = \underbrace{2(n) + 1}_{0\text{th \, term}} + \sum_{r=1}^n (2n+1) = (2n+1) + (2n+1)n = (2n+1)(n+1)

2.

\displaystyle \sum_{r=0}^n f(r) = \overbrace{f(0) + \underbrace{f(1) + f(2) + \cdots +f(n)}_{n \, \text{terms}}}^{(n+1) \, \text{terms}}

.

Since in this case you have

\sum_{r=0}^n (2n+1) = (2n+1)\sum_{r=0}^n 1

Then you have

f(r) = 1

being added together

n+1

times to get a total of

(2n+1)(1 + 1 + \cdots +1) = (2n+1)(n+1)

.

FP1 Summation Series when r=0

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