Ex.
1 + 2+3+...+ n = n(n+1)/2
Soln:
Let,
P(n): 1 + 2 + 3 + ... + n = n(n+1)/2
For n = 1,
LHS = 1
RHS = 1(1+1)/2 = 1x2/2 = 1
LHS = RHS
Let P(k) be true for some positive Integer i.e., k.
Let,
P(k): 1+2+3+...+ k = k(k+1)/2 be true ------------------------- (1)
we will prove P(k+1) is true,
Consider,
1+2+3+...+ k + k+1
= k(k+1)/2 + k+1 [Using eq. (1)]
= (k+1) (k+2) / 2
= (k+1) [(k+1) + 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 Number i.e., n.
(Question dang pawh hetianga chawh vek tur hi a ni ang)
2) 1+3+3^2+...+3^n-1 = (3^n - 1)/2
3) 1/1.2 + 1/2.3 + ... + 1/n(n+1) = n/n+1
4) 1.2 + 2.3 + ... + n(n+1) = n(n+1)(n+2)/3
1 + 2+3+...+ n = n(n+1)/2
Soln:
Let,
P(n): 1 + 2 + 3 + ... + n = n(n+1)/2
For n = 1,
LHS = 1
RHS = 1(1+1)/2 = 1x2/2 = 1
LHS = RHS
Let P(k) be true for some positive Integer i.e., k.
Let,
P(k): 1+2+3+...+ k = k(k+1)/2 be true ------------------------- (1)
we will prove P(k+1) is true,
Consider,
1+2+3+...+ k + k+1
= k(k+1)/2 + k+1 [Using eq. (1)]
= (k+1) (k+2) / 2
= (k+1) [(k+1) + 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 Number i.e., n.
(Question dang pawh hetianga chawh vek tur hi a ni ang)
2) 1+3+3^2+...+3^n-1 = (3^n - 1)/2
3) 1/1.2 + 1/2.3 + ... + 1/n(n+1) = n/n+1
4) 1.2 + 2.3 + ... + n(n+1) = n(n+1)(n+2)/3