Question -
Answer -
Let P (n) = 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4n-1)(4n+3) =n/3(4n+3)
Let us check for n = 1,
P (1): 1/3.7 = 1/(4.1-1)(4+3)
: 1/21 = 1/21
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): 1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k-1)(4k+3) =k/3(4k+3) …. (i)
So,
1/3.7 + 1/7.11 + 1/11.15 + … + 1/(4k-1)(4k+3) +1/(4k+3)(4k+7)
Substituting the value of P (k) we get,
= k/(4k+3) + 1/(4k+3)(4k+7)
= 1/(4k+3) [k(4k+7)+3] / [3(4k+7)]
= 1/(4k+3) [4k2 + 7k +3]/ [3(4k+7)]
= 1/(4k+3) [4k2 + 3k+4k+3] / [3(4k+7)]
= 1/(4k+3) [4k(k+1)+3(k+1)]/ [3(4k+7)]
= 1/(4k+3) [(4k+3)(k+1)] / [3(4k+7)]
= (k+1) / [3(4k+7)]
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.