Question -
Answer -
Let P (n) = 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3n-2)(3n+1) =n/3n+1
Let us check for n = 1,
P (1): 1/1.4 = 1/4
1/4 = 1/4
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) = 1/1.4 + 1/4.7 + 1/7.10 + … + 1/(3k-2)(3k+1) = k/3k+1… (i)
So,
[1/1.4 + 1/4.7 + 1/7.10 + … +1/(3k-2)(3k+1)]+ 1/(3k+1)(3k+4)
= k/(3k+1) + 1/(3k+1)(3k+4)
= 1/(3k+1) [k/1 + 1/(3k+4)]
= 1/(3k+1) [k(3k+4)+1]/(3k+4)
= 1/(3k+1) [3k2 + 4k + 1]/ (3k+4)
= 1/(3k+1) [3k2 + 3k+k+1]/(3k+4)
= [3k(k+1) + (k+1)] / [(3k+4) (3k+1)]
= [(3k+1)(k+1)] / [(3k+4) (3k+1)]
= (k+1) / (3k+4)
P (n) is true for n = k + 1
Hence, P (n) is true for all n ∈ N.