Stuck on AS Further Maths Series Question!

I wrote the series as an expression first which gave me:

n(n+1)(2n+1)/2 - n(n+1) -1

I then simplified this expression to n(n+1)(2n-1)/2 - 1.

I'm unsure on what steps I need to take to get to the answer, which is:
n(2n+3)(n-1)/2

See question below:
Any help would be appreciated!

(edited 6 years ago)

Reply 1

dotoswift

OP

7

Question:

Screen Shot 2018-03-01 at 13.52.42.png75.5KB

Original post by dotoswift

Question:

That's not entirely correct.

Note that

\displaystyle \sum_{r=1}^n 3r^2-2r-1 = 3\sum_{r=1}^n r^2 - 2 \sum_{r=1}^n r - \sum_{r=1}^n 1

Now note the last sum - it is not 1

Reply 3

dotoswift

OP

7

Original post by RDKGames

That's not entirely correct.

Note that

\displaystyle \sum_{r=1}^n 3r^2-2r-1 = 3\sum_{r=1}^n r^2 - 2 \sum_{r=1}^n r - \sum_{r=1}^n 1

Now note the last sum - it is not 1

Is this the first step then? :

n(n+1)(2n+1)/2 -n(n+1) - (3-2-1)

Original post by dotoswift

Is this the first step then? :

n(n+1)(2n+1)/2 -n(n+1) - (3-2-1)

Why? Where did 3-2-1 come from? Try to justify every move you make otherwise you end up writing rubbish without a reason.

The first two expressions are correct, so don't try to change those.

Summing up 1 an

n

number of times means

1+1+1+1+...+1

a total of

n

times. Hence what is

\displaystyle \sum_{r=1}^{n} 1

?

Reply 5

dotoswift

OP

7

Original post by RDKGames

Why? Where did 3-2-1 come from? Try to justify every move you make otherwise you end up writing rubbish without a reason.

The first two expressions are correct, so don't try to change those.

Summing up 1 an