Question

Question 10 Prove the following by using the principle of mathematical induction for all n∈N: 1/2.5 + 1/5.8 + 1/8.11 + ... + 1/(3n-1)(3n+2) = n / (6n+4)

Class X1 - Maths -Principle of Mathematical Induction Page 94

Answer 1

1/2.5 + 1/5.8 + 1/8.11 + .......+ 1/(3n-1)(3n+2) = n/(6n+4)

Let P(n): 1/2.5 + 1/5.8 + 1/8.11 + .......+ 1/(3n-1)(3n+2) = n/(6n+4)

step1:- for n = 1
P(1) = 1/2.5 = 1/(6+4) = 1/10 = 1/2.5
which is true.

step2:- for n = k
P(K): 1/2.5 + 1/5.8 + 1/8.11 + .......+ 1/(3k-1)(3k+2) = k/(6k+4) _______(1)

step3:- for n = (k+1)
P(k+1) = 1/2.5 + 1/5.8 + 1/8.11 + .......+ 1/(3k+2)(3k+5) = (k+1)/(6k+10)

from equation (1)
P(K): 1/2.5 + 1/5.8 + 1/8.11 + .......+ 1/(3k-1)(3k+2) = k/(6k+4)

add ' 1/(3k+2)(3k+5)' both sides,

1/2.5 + 1/5.8 + 1/8.11 + .......+ 1/(3k-1)(3k+2) + 1/(3k+2)(3k+5) = k/(6k+4) + 1/(3k+2)(3k+5)
= k/2(3k+2) + 1/(3k+2)(3k+5)
= {k(3k+5) + 2}/2(3k+2)(3k+5)
={3k² + 5k +2}/2(3k+2)(3k+5)
= (k+1)(3k+2)/2(3k+2)(3k+5)
=[(k+1)/{6(k+1) +4}]
P(k+1) is true when P(k) is true . from the principle of Mathematical induction, statement is true for all natural numbers.