Doing my A level maths homework on sequences and series and am getting super stuck on this one questions.
Any help / advice would be much appreciated, and anyone who provides a solution gets a cookie and karma!
Ok,
"Prove that the sum of the first 2n multiples of 4 is 4n(2n+1)"
The formula you need to use is Sn = n/2 (2a + (n-1) d)
Where a is the first term, d is the difference and n is the nth term.
Heres an example at one of my attempts to give you the jist of it...
(n x 4) 4 + 8 + 12 + 16 +....
(2n x 4) 8 + 16 + 24 + 32 +...
a = 8, d = 8, n = 8n
Sn = n/2 (2a + (n-1) d)
S8n = 4n (16 + 64n - 8 )
S8n = 4n (64n + 8)
S8n = 4n (8n + 1)
But we need to get 4n(2n+1),
Thanks for the help!
Any help / advice would be much appreciated, and anyone who provides a solution gets a cookie and karma!
Ok,
"Prove that the sum of the first 2n multiples of 4 is 4n(2n+1)"
The formula you need to use is Sn = n/2 (2a + (n-1) d)
Where a is the first term, d is the difference and n is the nth term.
Heres an example at one of my attempts to give you the jist of it...
(n x 4) 4 + 8 + 12 + 16 +....
(2n x 4) 8 + 16 + 24 + 32 +...
a = 8, d = 8, n = 8n
Sn = n/2 (2a + (n-1) d)
S8n = 4n (16 + 64n - 8 )
S8n = 4n (64n + 8)
S8n = 4n (8n + 1)
But we need to get 4n(2n+1),
Thanks for the help!