Question

Question 16 Prove the following by using the principle of mathematical induction for all n∈N: 1/1.4 + 1/4.7 + 1/7.10 + ... + 1/(3n-2)(3n+1) = n/(3n+1)

Class X1 - Maths -Principle of Mathematical Induction Page 95

Answer 1

1/1.4 + 1/4.7 + 1/7.10 +.....+ 1/(3n-2)(3n+1) = n/(3n+1)
Let p(n): 1/1.4 + 1/4.7 + 1/7.10 +.....+ 1/(3n-2)(3n+1) = n/(3n+1)

step1:- for n = 1
P(1) : 1/1.4 = 1/(3×1+1) = 1/4 = 1/1.4
it's true.

step2:- for n = k
p(k): 1/1.4 + 1/4.7 + 1/7.10 +.....+ 1/(3k-2)(3k+1) = k/(3k+1) ------(1)

step3:-
from eqn (1)
1/1.4 + 1/4.7 + 1/7.10 +.....+ 1/(3k-2)(3k+1) = k/(3k+1)
add 1/(3k+1)(3k+4) both sides,
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)
= {k(3k+4)+1}/(3k+1)(3k+4)
= {3k² + 4k+1}/(3k+1)(3k+4)
= (3k+1)(k+1)/(3k+1)(3k+4)
= (k+1)/[3(k+1)+1]
hence,
P(k+1) is true when p(k) is true . from the principle of mathematical induction , statement is true for all natural numbers.