Question -
Answer -
Let P (n) = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2n+1)(2n+3) =n/3(2n+3)
Let us check for n = 1,
P (1): 1/3.5 = 1/3(2.1+3)
: 1/15 = 1/15
P (n) is true for n = 1.
Now, let us check for P (n) is true for n = k, and have toprove that P (k + 1) is true.
P (k) = 1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) =k/3(2k+3) … (i)
So,
1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) +1/[2(k+1)+1][2(k+1)+3]
1/3.5 + 1/5.7 + 1/7.9 + … + 1/(2k+1)(2k+3) + 1/(2k+3)(2k+5)
Now substituting the value of P (k) we get,
= k/3(2k+3) + 1/(2k+3)(2k+5)
= [k(2k+5)+3] / [3(2k+3)(2k+5)]
= (k+1) / [3(2(k+1)+3)]
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.