Show that the sum from 1 to n of 1/(2n+1)(2n-1) is equal to n/(2n+1) by Induction

First we check that this is true for n=1: S1 = 1/(1x3)  which is equal to n/(2n+1) for n=1 therefore Sn = n/(2n+1) is true for n = 1. Next assume that it is true for n=k. Sk  = k/(2k+1). Now using this assumption we check that it is true for n=k+1: Sk+1 = Sk+ 1/(2(k+1) - 1)(2(k+1)+1). Rearranging this and substituting in k/(2k+1) for Sk we get Sk+1 = (k+1)/(2k+3) which is consistent with the original formula. Therefore we can say that since Sn = n/(2n+1) is true for n=1 and whenever it is true for n=k it is also true for n=k+1, it is true for all integer values of n larger than or equal to 1.

JF

Related Further Mathematics A Level answers

All answers ▸

Find the general solution of the differential equation d^2y/dx^2 - 2(dy/dx) = 26sin(3x)


using an integrating factor, find the general solution of the differential equation dy/dx +y(tanx)=tan^3(x)sec(x)


Differentiate w.r.t x the expression arccos(x).


Convert the general complex number z=x+iy to modulus-argument form.