Find the sum up to n terms in 1/1*3*5+1/3*5*7+1/5*7*9+.............. using any of the standard four methods.?

Answer 1

the answer is 0

Answer 2

The best way I could find is to set up a difference equation. Not sure what the "four standard methods" are, but this one clearly works.

Let f(n) = 1/((n)(n+2)). Then f(n) - f(n + 2) = 1/n(n+2) - 1/(n+2)(n+4) = 4/n(n+2)(n+4).

Thus letting g(n) = f(n)/4 we get the term we want. Then

1/(1*3*5) = g(1) - g(3)
1/(3*5*7) = g(3) - g(5)
and so on.

summing up to infinity gives a telescoping sum that gives g(1) - g(infinity). g(infinity) = 0, so the total sum is g(1) = (1/4)(1/1)(1/3) = 1/12.

Answer 3

This is a complicated problem in telescoping series. The expression

1/[(2k-1)(2k+1)(2k+3)] breaks into the expression 1/8 times

1/(2k-1) - 2/(2k+1) + 1/(2k+3).

Since 1 - 2 + 1 = 0, all the middle terms cancel, leaving just a few in the beginning and in the end. The answer is 1/8 times

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