Question

Prove using the principle of mathematical induction ​

Answer 1

Question

Prove using principle of mathematical induction

1*3 + 3*5 + 5*7 +.....+ (2n - 1)(2n + 1) = $\sf\frac{n(4n^2+6n-1)}{3}$

Solution

First of all we will check whether the expression is true for P(1), that is, n = 1

1*3 = $\sf\frac{1(4(1)^2+6(1)-1)}{3}$

3 = $\sf\frac{9}{3}$

3 = 3

LHS = RHS

Hence, the given expression is true for P(1)

Now we will assume that given expression is true for P(k) for some positive integer k. That is,

1*3 + 3*5 ....... + (2k - 1)(2k + 1) = $\sf\frac{k(4k^2+6k-1)}{3}$

Now we will check if it is true for integer (k + 1) too.

1*3 + 3*5 .......... + (2k - 1)(2k + 1) + (2(k + 1) - 1)(2(k + 1) + 1)

⇒ 1*3 + 3*5 .......... + (2k - 1)(2k + 1) + (2k + 2 - 1)(2k + 2 + 1)

⇒ $\sf\frac{k(4k^2+6k-1)}{3}$ + (2k + 1)(2k + 3)

(Since 1*3 + 3*5 ....... + (2k - 1)(2k + 1) = $\sf\frac{k(4k^2+6k-1)}{3}$ )

$\sf\frac{4k^3 + 6k^2 - k}{3}$ + (4k^2 + 8k + 3)

⇒ $\sf\frac{4k^3+6k^2-k+12k^2+24k+9}{3}$

⇒ $\sf\frac{4k^3+18k^2+23k+9}{3}$

Now, the given expression 4k³ + 18k² + 23k + 9 becomes 0 for k = -1. Hence, (k + 1) is a factor of 4k³ + 18k² + 23k + 9

$\sf\frac{4k^3+18k^2+23k+9}{3}$

Take out (k + 1) as factor

⇒ $\sf\frac{(k+1)(4k^2+ 14k + 9)}{3}$

⇒ $\sf\frac{(k+1)(4k^2+ 8k + 4 + 6k + 5)}{3}$

⇒ $\sf\frac{(k+1)(4(k^2+ 2k + 1) + 6k + 5)}{3}$

⇒ $\sf\frac{(k+1)(4(k + 1)^2 + 6k + 6 - 1)}{3}$

⇒ $\sf\frac{(k+1)(4(k + 1)^2 + 6(k + 1) - 1)}{3}$

Hence,

1*3 + 3*5 ..........+(2k - 1)(2k + 1) + (2(k + 1) - 1)(2(k + 1) + 1) = $\sf\frac{(k+1)(4(k + 1)^2 + 6(k + 1) - 1}{3}$

Hence, P(k + 1) is true whenever P(k) is true. Hence, by the principle of mathematical induction,

1*3 + 3*5 + 5*7 +.....+ (2n - 1)(2n + 1) = $\sf\frac{n(4n^2+6n-1)}{3}$

is true for all natural values of n

Answer 2

$\large{\underline{\bf{Proof:-}}}$

let P(n): 1.3+3.5+5.7+.....+(2n-1)(2n+1) = n(4n²+6n-1)/3

Now,

P(1) = 1.3= 3= 1(4.1²+6.1-1)/3 = 4+6-2/3

= 9/3 = 3

which is true.

Let P(k) be true for some positive integer k:

1.3+3.5+5.7+......+(2k-1)(2k+1)= k(4k²+6k-1)/3 ......(1)

Now, we prove P(k+1)

→ (1.3+3.5+5.7+.....+(2k-1)(2k+1)+{2(k+1)-1}{2(k+1)+1}

→ k(4k²+6k-1)/3+(2k+1-1)(2k+2+1)

→ k(4k²+6k-1)/3+(4k²+8k+3)

→ 4k²+6k²-k+12k²+24k+9/3

→ 4k³+14k²+9k+4k²+14k+9/3

→ k(4k²+14k+8)+1(4k²-14+9)/3

→ (k+1)(4k²+14k+9)/3

→ (k+1){4(k+1)№+6(k+1)-1}/3

•°• Thus P(k+1) is true.

Hence by PMI , statement P(n) is true for all natural no. n.