The Student Room Group
Uni students - Complete our survey >>
Uni students - Complete our survey >>
Uni students - Complete our survey >>

FP1 Series

Use standard results for ∑r=1nr2\displaystyle\sum_{r=1}^n r^2 and ∑r=1nr\displaystyle\sum_{r=1}^n r to show that

∑r=0n(r2−2r+2n+1)=16(n+1)(n+a)(bn+c)\displaystyle\sum_{r=0}^n (r^2 -2r + 2n + 1) = \frac{1}{6}(n+1)(n+a)(bn+c)
For all integers n≥0n \geq 0 where a, b and c are constant integers to be found.
I split up the series into:

∑r=0nr2−2∑r=0nr+2∑r=0nn+∑r=0n1\displaystyle\sum_{r=0}^n r^2 - 2\displaystyle\sum_{r=0}^n r + 2\displaystyle\sum_{r=0}^n n + \displaystyle\sum_{r=0}^n 1

I can work out the first two terms to get:

16n(n+1)(2n+1)−n(n+1)+2∑r=0nn+∑r=0n1\frac{1}{6}n(n+1)(2n+1) - n(n+1) + 2\displaystyle\sum_{r=0}^n n + \displaystyle\sum_{r=0}^n 1

But I'm not quite sure how to work out:

2∑r=0nn+∑r=0n12\displaystyle\sum_{r=0}^n n + \displaystyle\sum_{r=0}^n 1
Reply 1
Avatar for Ayman!
Ayman!
15
Original post by The_Big_E
Use standard results for ∑r=1nr2\displaystyle\sum_{r=1}^n r^2 and ∑r=1nr\displaystyle\sum_{r=1}^n r to show that

∑r=0n(r2−2r+2n+1)=16(n+1)(n+a)(bn+c)\displaystyle\sum_{r=0}^n (r^2 -2r + 2n + 1) = \frac{1}{6}(n+1)(n+a)(bn+c)
For all integers n≥0n \geq 0 where a, b and c are constant integers to be found.
I split up the series into:

∑r=0nr2−2∑r=0nr+2∑r=0nn+∑r=0n1\displaystyle\sum_{r=0}^n r^2 - 2\displaystyle\sum_{r=0}^n r + 2\displaystyle\sum_{r=0}^n n + \displaystyle\sum_{r=0}^n 1

I can work out the first two terms to get:

16n(n+1)(2n+1)−n(n+1)+2∑r=0nn+∑r=0n1\frac{1}{6}n(n+1)(2n+1) - n(n+1) + 2\displaystyle\sum_{r=0}^n n + \displaystyle\sum_{r=0}^n 1

But I'm not quite sure how to work out:

2∑r=0nn+∑r=0n12\displaystyle\sum_{r=0}^n n + \displaystyle\sum_{r=0}^n 1


Do you know the formulae for the two sums you can't break down?
Original post by The_Big_E


2∑r=0nn+∑r=0n12\displaystyle\sum_{r=0}^n n + \displaystyle\sum_{r=0}^n 1


Hint: When you have a constant there:

∑r=0na=a+a+⋯+a+a⏟n+1 times=...?\displaystyle\sum_{r=0}^n a=\underbrace{a+a+\cdots +a+a}_{\text{n+1 times}}=...?
Original post by The_Big_E

But I'm not quite sure how to work out:

2∑r=0nn+∑r=0n12\displaystyle\sum_{r=0}^n n + \displaystyle\sum_{r=0}^n 1


Remember that n is a constant, so

∑r=0nn=n×∑r=0n1\displaystyle\sum_{r=0}^n n = n \times \sum_{r=0}^n 1

So we're now down to working out what

∑r=0n1\displaystyle\sum_{r=0}^n 1

is. Whenever you have a puzzling sum, write it down explicitly. This one is simply

1+1+…+1 1 +1 + \ldots + 1

for as many terms as there are in the summation. Got it now?
Reply 4
Avatar for The_Big_E
The_Big_E
OP
11
Original post by Gregorius
Remember that n is a constant, so

∑r=0nn=n×∑r=0n1\displaystyle\sum_{r=0}^n n = n \times \sum_{r=0}^n 1

So we're now down to working out what

∑r=0n1\displaystyle\sum_{r=0}^n 1

is. Whenever you have a puzzling sum, write it down explicitly. This one is simply

1+1+…+1 1 +1 + \ldots + 1

for as many terms as there are in the summation. Got it now?


I get the idea, but i'm not quite sure what to do when r = 0. For ∑r=1n1\displaystyle\sum_{r=1}^n 1, it would just be n right? But what about when r = 0?
Original post by The_Big_E
I get the idea, but i'm not quite sure what to do when r = 0. For ∑r=1n1\displaystyle\sum_{r=1}^n 1, it would just be n right? But what about when r = 0?


r is just a "dummy" variable that does the counting in the sum. So here it just gives you one more term in the sum, so the answer is n+1.

Quick Reply

Related discussions

Latest

Trending

Trending

Articles for you