NCERT Solutions
Class 11 Maths
Principle of Mathematical Induction
Ex.4.1 Q.11
Prove the following by using the principle of mathematical induction for all n є N:
+ 
+ 
+ …………...+ [1 ÷ {n (n + 1) (n + 2)}] = {n (n + 3)} ÷ {4(n + 1) (n + 2)}
Let the given statement be P(n), i.e.,
P(n): 
+ 
+ 
+ …………...+ 1 ÷ {n (n + 1) (n + 2)} = {n (n + 3)} ÷ {4(n + 1) (n + 2)}
For n = 1, we have
P (1): = {1(1 + 3)} ÷ {4(1 + 1) (1 + 2)} = (1.4) ÷ (4.2.3) = , which is true.
Let P(k) be true for some positive integer k, i.e.,
+ + + …………...+ 1 ÷ {k (k + 1) (k + 2)} = {k (k + 3)} ÷ {4(k + 1) (k + 2)}
We shall now prove that P (k + 1) is true.
Consider
+ + + …………...+ [1 ÷ {k (k + 1) (k + 2)}] + [1 ÷ {(k + 1) (k + 2) (k + 3)}]
= {k (k + 3)} {4(k + 1) (k + 2)} + [1 ÷ {(k + 1) (k + 2) (k + 3)}]
[From equation 1]
= [1 ÷ {(k + 1) (k + 2)}]{k (k + 3) ÷ 4 + 1 ÷ (k + 3)}
= [1 ÷ {(k + 1) (k + 2)}][{k (k + 3)2 + 4} ÷ (k + 3)]
= [1 ÷ {(k + 1) (k + 2)}][{k (k2 + 6k + 9) + 4} ÷ (k + 3)]
= [1 ÷ {(k + 1) (k + 2)}][{k3 + 6k2 + 9k + 4} ÷ (k + 3)]
= [1 ÷ {(k + 1) (k + 2)}][{k3 + 2k2 + k + 4k2 + 8k + 4} ÷ (k + 3)]
= [1 ÷ {(k + 1) (k + 2)}][{k (k2 + 2k + 1) + 4(k2 + 2k + 1)} ÷ (k + 3)]
= [1 ÷ {(k + 1) (k + 2)}][{k (k + 1)2 + 4(k + 1)2} ÷ (k + 3)]
= [1 ÷ {(k + 1) (k + 2)}][{(k + 4) (k + 1)2} ÷ (k + 3)]
= {(k + 4) (k + 1)2} ÷ {4(k + 1) (k + 2) (k + 3)}
= {(k + 4) (k + 1)} ÷ {4(k + 2) (k + 3)}
= {(k + 1) (k + 1 + 3)} ÷ {4(k + 1 + 1) (k + 1 + 2)}
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.