Question

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

Class X1 - Maths -Principle of Mathematical Induction Page 94

Answer 1

1 + 1/(1 + 2) + 1/(1 + 2 + 3) + ...........+1/(1 + 2 + 3 ....n) = 2n/(n+1)

Let p(n): 1 + 1/(1 + 2) + 1/(1 + 2 + 3) +......1/(1 + 2 + 3 +...n) = 2n/(n+1)

step1:- for n=1
P(1) : 1 = 2×1/(1+1) = 1
It's true .

step2:- for n = k
P(k): 1 + 1/(1 + 2) + 1/(1 + 2 + 3) + .......1/(1 + 2 + 3 ....k) = 2k/(k+1)--------(1)

step3:- for n= k+1
From eqn (1)
1 + 1/(1 + 2) + 1/(1 + 2 + 3) + ........+ 1/(1 + 2 + 3....+k) = 2k/(k+1)
add ( k+1) both sides,
1 + 1/(1+2) + 1/(1+2+3) + ......+1/(1 + 2 + ....+k) + 1/{1 + 2+3......+k+(k+1)} = 2k/(k+1) + 1/{1 +2 +3 +....k+(k+1)}
= 2k/(k+1) + 2/(k+1)(k+2)
= {2k(k+2)+2}/(k+1)(k+2)
= {2k² + 4k + 8}/(k+1)(k+2)
= 2(k²+4k+4)/(k+1)(k+2)
=2(k+1)²/(k+1)(k+2)
=2(k+1)/(k+2)
= 2(k+1)/[(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.