Answer to Question #198624 in Calculus for Moe

Question

3) Prove by induction that for all natural numbers n:

(d) 1/(1)(3) + 1/(3)(5) + · · · +1/(2n − 1)(2n + 1) = n/(2n + 1)

Answer 1

Step 1: Given data

P(n) : $\dfrac{1}{1.3}+\dfrac{1}{3.5}+−−−−−+\dfrac{1}{(2n−1)(2n+1)}=\dfrac{n}{(2n+1)}$

Step 2: To prove, $P(1)$ is true

 $LHS=\dfrac{1}{1.3}=\dfrac{1}{3}$

$RHS=\dfrac{1}{2×1+1}=\dfrac{1}{3}$

Thus, $P(1)$ is true.

Step 3: Assume that the statement is true for $n = k$ , where k is some positive integer.

$\dfrac{1}{1.3}+\dfrac{1}{3.5}+−−−−−+\dfrac{1}{(2k−1)(2k+1)}=\dfrac{k}{(2k+1)}$

 

Step 4:  For $n=k+1$  

$LHS= \dfrac{1}{1.3}+\dfrac{1}{3.5}+−−−−+\dfrac{1}{(2k−1)(2k+1)}+\dfrac{1}{(2(k+1)−1)(2(k+1)+1)}$

$=\dfrac{k}{(2k+1)}+\dfrac{1}{(2(k+1)−1)(2(k+1)+1)}$     

 $=\dfrac{k}{(2k+1)}+\dfrac{1}{(2k+1)(2k+3)}$

$=\dfrac{2k^2+3k+1}{(2k+1)(2k+3)}$

$=\dfrac{(2k+1)(k+1)}{(2k+1)(2k+3)}$

$LHS=\dfrac{(k+1)}{(2k+3)}$

  

Step 5:

$RHS=\dfrac{(k+1)}{(2(k+1) + 1)}$

$=\dfrac{k+1} { 2k + 3}$

⇒LHS=RHS