Question

please solve ques 13 as soon as possible

Answer 1

Step-by-step explanation:

It can be solved using the method of proof of induction which involves the following steps:

⇒ Prove true for some value, say n = 1

⇒ Assume the result is true for n = k.

⇒ Prove true for n = k + 1.

(i) n = 1

LHS = (1 + 3/1) = 1 + 3 = 4

RHS = (1 + 1)² = 4

Result is true for n = 1.

(ii) n = k:

(1 + 3)(1 + 5/4)(1 + 7/9)...(1 + (2k+1)/k²) = (k + 1)²

(iii) n = k + 1:

LHS:

$\Longrightarrow(1 + \frac{3}{1})(1 + \frac{5}{4})(1 + \frac{7}{9})....(1+\frac{2(k+1)+1}{(k+1)^2})$

$\Longrightarrow(1+\frac{3}{1})(1+\frac{5}{4})....(1+\frac{2k+1}{k^2})(1+\frac{2(k+1)+1}{(k+1)^2})$

$\Longrightarrow(k+1)^2(1+\frac{2(k+1)+1}{(k+1)^2}) [\because \; from \; (ii)]$

$\Longrightarrow(k+1)^2[\frac{(k+1)^2+(2k+3)}{(k+1)^2}]$

$\Longrightarrow(k+1)^2+(2k+3)$

$\Longrightarrow k^2+1+2k+2k+3$

$\Longrightarrow k^2 + 4k + 4$

RHS:

$\Longrightarrow(k+1+1)^2$

$\Longrightarrow(k+2)^2$

$\Longrightarrow(k^2+4+4k)$

LHS = RHS

∴ P(k + 1) is true whenever p(k) is true.

Hope it helps!