The Student Room Group
Reply 1
futureaussiecto
Erm- how do you do these? Ive forgotten and the book is useless.......

For example im doing

Ex 2A 4)

In each case use the identity given to find the sum to n terms of the given series.

Identity
r²(r+1)-(r-1)²(r)≡3r²-r

Series

n∑r=1 r(3r-1)

(1)=3-1
(2)=12-2
(3)=27-3
(4)=48-4
(5)=75-5
(6)=108-6

im not too sure what im doing now??


Notice that 3r²-r=r(3r-1) which is the series you want to sum.
So ∑r(3r-1)=∑r²(r+1)-(r-1)²(r)
(1²)(2)-(0²)(1)
(2²)(3)-(1²)(2)
(3²)(4)-(2²)(3)
.
.
(n-1)²n-(n-2)² (n-1)
n²(n+1)-(n-1)²n

You should notice that any element on the left of our term will cancel with the part on the right of the term below.
So on adding them we are left with n²(n+1) which is the sum of the series.
Reply 2
futureaussiecto
Erm- how do you do these? Ive forgotten and the book is useless.......

For example im doing

Ex 2A 4)

In each case use the identity given to find the sum to n terms of the given series.

Identity
r²(r+1)-(r-1)²(r)≡3r²-r

Series

n∑r=1 r(3r-1)

(1)=3-1
(2)=12-2
(3)=27-3
(4)=48-4
(5)=75-5
(6)=108-6

im not too sure what im doing now??


Use the left-hand side of your identity:
(1)2-0
(2)12-2
(3)36-12
(4)80-36
(n)n²(n+1)...

cancel diagonally.
leaves n²(n+1)-0
Gaz031
(n-1)²n-(n-2)² (n-1)
n²(n+1)-(n-1)²n+1)


can i just add up these last two lines and divide by 2?
Reply 4
futureaussiecto
can i just add up these last two lines and divide by 2?

No - these are the final two terms of your series. You need to add all the terms listed.
cheers guys! just did number 5 and got it right!!

for some reason i do find fp1 easier than c4!!

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