Find the sum to n terms of 1/(1.2.3) + 3/(2.3.4) + 5/(3.4.5) + 7/(4.5.6)

Answer 1

I get

n(3n+1)
---------------
4(n^2+3n+2)

How did you do it?

Answer 2

Yes,your answer is correct. Can you please show your workings?

Answer 3

use induction. Let the sum be f(n)

for k=1, it is clear that

1/(1*2*3) = 1(4)/(4*6)

Now assume true for k=n and see what happens for k=n+1

1/(1*2*3) + ... + (2n-1)/(n(n+1)(n+2)) + (2(n+1)-1)/((n+1)(n+2)(n+3))
= n(3n+1)/(4(n+1)(n+2)) + (2n+1)/(n+1)(n+2)(n+3)
=

n(3n+1)(n+3) + 4(2n+1)
------------------------
4(n+1)(n+2)(n+3)
=
3n^3+10n^2+3n+8n+4
-----------------------
4(n+1)(n+2)(n+3)
=
3n^3+10n^2+11n+4
----------------------
4(n+1)(n+2)(n+3)
=
(n+1)^2(3n+4)
------------------
4(n+1)(n+2)(n+3)
=
(n+1)(3(n+1)+1) / 4(n+1)(n+2)

which is f(n+1)

Answer 4

The sum of the first "n" termsof a series 2^(n+3) -8.Find the 7th term.