Question

class 11 principal of mathematical induction ​

Answer 1

Let ,

$\sf \star \: \: P(n) = 1 + 4 + 7 + ... + (3n + 2) = \frac{n(3n - 1)}{2} - - - (i)$

LHS of eq (i) is

$\sf \mapsto(3 × 1 - 2) \\ \\ \sf \mapsto</p><p>1$

and

RHS of eq (i) is

$\sf \mapsto \frac{1(3 \times 1 - 1)}{2} \\ \\ \sf \mapsto1$

$\sf \therefore \underline{LHS = RHS}$

Therefore , P(1) is true

Assume P(k) is true , then

$\sf \star \: \: P(k) = 1 + 4 + 7 + ... + (3k + 2) = \frac{k(3k - 1)}{2}$

For n = k + 1

$\sf \mapsto P(k + 1) = 1 + 4 + 7 + ... + (3k - 2) + \{3(k + 1) - 2 \}= \frac{(k + 1) \{3(k + 1) - 1 \}}{2} \\ \\ \sf \mapsto P(k + 1) = P(k ) + 3k + 1= \frac{(k + 1)(3k + 2)}{2} \\ \\ \sf \mapsto P(k + 1) = P(k ) + 3k + 1 = \frac{3 {k}^{2} + 2k + 3k + 2}{2} \\ \\ \sf \mapsto P(k + 1) = P(k ) + 3k + 1 = \frac{3 {k}^{2} + 5k + 2}{2}$

LHS :

$\sf \mapsto P(k + 1) = P(k ) + 3k + 1 \\ \\ \sf \mapsto P(k + 1) = \frac{k(3k - 1)}{2} + 3k + 1\\ \\ \sf \mapsto P(k + 1) = \frac{3 {k}^{2} - k + 6k + 2}{2} \\ \\ \sf \mapsto P(k + 1) = \frac{3 {k}^{2} + 5k + 2}{2}$

$\sf \therefore \underline{LHS = RHS}$

Therefore , P(n) is true for all natural number