NCERT Solutions
Class 11 Maths
Principle of Mathematical Induction
Ex.4.1 Q.10
Prove the following by using the principle of mathematical induction for all n є N:
+ 
+ 
+ …………. + [1 ÷ {(3n - 1) (3n + 1)}] = [n ÷ (6n + 4)]
Let the given statement be P(n), i.e.,
P(n): 
+ 
+ 
+ …………. + [1 ÷ {(3n - 1) (3n + 1)}] = [n ÷ (6n + 4)]
For n = 1, we have
P (1): = 
= 1 ÷ (6.1 + 4) = , which is true.
Let P(k) be true for some positive integer k, i.e.,
1 ÷ {(3k - 1) (3k + 1)} = {k ÷ (6k + 4)} …………… (1)
We shall now prove that P (k + 1) is true.
Consider
+ + + …………. + 1 ÷ {(3k - 1) (3k + 1)} + 1 ÷ [{3(k + 1) – 1} {3(k + 1) + 2}]
= {k ÷ (6k + 4)} + [1 ÷ [{3(k + 1) – 1} {3(k + 1) + 2}]]
[using equation 1]
= (k ÷ (6k + 4)) + 1 ÷ {(3k + 3 – 1) (3k + 3 + 2)}
= (k ÷ (6k + 4)) + [1 ÷ {(3k + 2) (3k + 5)}]
= (k ÷ {2(3k + 2)} + [1 ÷ {(3k + 2) (3k + 5)}]
= 1 ÷ (3k + 2) { 
+ 
}
= 
[{k (3k + 5) + 2} ÷ {2(3k + 5)}]
= [{(3k2 + 5k + 2} ÷ {6k + 10}]
= [{(3k + 2) (k + 1)} ÷ {6k + 10}]
= (k + 1) ÷ (6k + 10)
= (k + 1) ÷ {6(k + 1) + 4}
Thus, P (k + 1) is true whenever P(k) is true.
Hence, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e., N.