Question

Prove the mathematical induction

${1}^{2} + {3}^{2} + {5}^{2} + ... + {(2n - 1)}^{2} = \frac{1}{3} n (2n - 1)(2n + 1)$
​

Answer 1

$\large\boxed{\textsf{\textbf{\pink{Question\::-}}}}$

• Prove by Mathematical induction that

${1}^{2} + {3}^{2} + {5}^{2} ....(2n - 1) =$

$\frac{n(2n - 1)(2n + 1)}{3} ∀n∈N$

$\large\boxed{\textsf{\textbf{\pink{Answer\::-}}}}$

To prove:

${1}^{2} + {3}^{2} + {5}^{2} ... + (2n - 1) {}^{2} =$

$\frac{n(2n - 1)(2n + 1)}{3} ∀n∈N$

Proof:

$P \: (n) = {1}^{2} + {3}^{2} + {5}^{2} ... + (2n - 1) {}^{2} =$

$\frac{n(2n - 1)(2n + 1)}{3}$

$P \: (1):(2×1−1) {}^{2} = \frac{1(2 - 1)(2 + 1)}{3}$

$⇒(1) {}^{2} = 1 = \frac{1 \times 1 \times 3}{3} = 1$

∴ L.H.S=R.H.S (Proved)

∴P(1) is true

Now, let P(m) is true.

Then,

$P(m)=1 {}^{2} + {3}^{2} + {5}^{2} ... + (2m - 1) {}^{2} =$

$\frac{m(2m - 1)(2m + 1)}{3}$

Now, we have to prove that P(m+1) is also true.

$P(m+1)=1 {}^{2} + {3}^{2} + {5}^{2} ... + (2m + 1) {}^{2} +$

$[2(m+1)−1] {}^{2}$

$=P(m)+(2m+2−1) {}^{2}$

$=P(m)+(2m+1) {}^{2}$

$= \frac{m(2m - 1)(2m + 1)}{3} + (2m + 1) {}^{2}$

$= \frac{m(2m - 1)(2m + 1) + 3(2m + 1) {}^{2} }{3}$

$= \frac{</p><p>(2m+1)[m(2m−1)+3(2m+1)]}{3}$

$= \frac{(2m+1)[2m {}^{2} - m + 6m + 3] }{3}$

$= \frac{(2m + 1)[2m{}^{2} + 5m +3]}{3}$

$= \frac{(2m + 1)[2m {}^{2} + 2m + 3m + 3]}{3}$

$= \frac{(2m+1)[2m(m+1)+3(m+1)]}{3}$

$= \frac{(2m+1)(2m+3)(m+1)}{3}$

$= \frac{(m+1)[2(m+1)+1][2(m+1)−1]</p><p>}{3}$

∴p(m+1) is also true (Proved)

Answer 2

Answer:

\large\boxed{\textsf{\textbf{\pink{Question\::-}}}}Question:-

Prove by Mathematical induction that

{1}^{2} + {3}^{2} + {5}^{2} ....(2n - 1) =12+32+52....(2n−1)=

\frac{n(2n - 1)(2n + 1)}{3} ∀n∈N3n(2n−1)(2n+1)∀n∈N

\large\boxed{\textsf{\textbf{\pink{Answer\::-}}}}Answer:-

To prove:

{1}^{2} + {3}^{2} + {5}^{2} ... + (2n - 1) {}^{2} =12+32+52...+(2n−1)2=

\frac{n(2n - 1)(2n + 1)}{3} ∀n∈N3n(2n−1)(2n+1)∀n∈N

Proof:

P \: (n) = {1}^{2} + {3}^{2} + {5}^{2} ... + (2n - 1) {}^{2} =P(n)=12+32+52...+(2n−1)2=

\frac{n(2n - 1)(2n + 1)}{3}3n(2n−1)(2n+1)

P \: (1):(2×1−1) {}^{2} = \frac{1(2 - 1)(2 + 1)}{3}P(1):(2×1−1)2=31(2−1)(2+1)

⇒(1) {}^{2} = 1 = \frac{1 \times 1 \times 3}{3} = 1⇒(1)2=1=31×1×3=1

∴ L.H.S=R.H.S (Proved)

∴P(1) is true

Now, let P(m) is true.

Then,

P(m)=1 {}^{2} + {3}^{2} + {5}^{2} ... + (2m - 1) {}^{2} =P(m)=12+32+52...+(2m−1)2=

\frac{m(2m - 1)(2m + 1)}{3}3m(2m−1)(2m+1)

Now, we have to prove that P(m+1) is also true.

P(m+1)=1 {}^{2} + {3}^{2} + {5}^{2} ... + (2m + 1) {}^{2} +P(m+1)=12+32+52...+(2m+1)2+

[2(m+1)−1] {}^{2}[2(m+1)−1]2

=P(m)+(2m+2−1) {}^{2}=P(m)+(2m+2−1)2

=P(m)+(2m+1) {}^{2}=P(m)+(2m+1)2

= \frac{m(2m - 1)(2m + 1)}{3} + (2m + 1) {}^{2}=3m(2m−1)(2m+1)+(2m+1)2

= \frac{m(2m - 1)(2m + 1) + 3(2m + 1) {}^{2} }{3}=3m(2m−1)(2m+1)+3(2m+1)2

= \frac{ < /p > < p > (2m+1)[m(2m−1)+3(2m+1)]}{3}=3</p><p>(2m+1)[m(2m−1)+3(2m+1)]

= \frac{(2m+1)[2m {}^{2} - m + 6m + 3] }{3}=3(2m+1)[2m2−m+6m+3]

= \frac{(2m + 1)[2m{}^{2} + 5m +3]}{3}=3(2m+1)[2m2+5m+3]

= \frac{(2m + 1)[2m {}^{2} + 2m + 3m + 3]}{3}=3(2m+1)[2m2+2m+3m+3]

= \frac{(2m+1)[2m(m+1)+3(m+1)]}{3}=3(2m+1)[2m(m+1)+3(m+1)]

= \frac{(2m+1)(2m+3)(m+1)}{3}=3(2m+1)(2m+3)(m+1)

= \frac{(m+1)[2(m+1)+1][2(m+1)−1] < /p > < p > }{3}=