Question

1.3 + 3.5 + 5.7 + ... + (2n − 1) (2n + 1) =\[\frac{n(4 n^2 + 6n - 1)}{3}\]

 

Answer 1

Let P(n) be the given statement.
Now,

\[P(n) = 1 . 3 + 3 . 5 + 5 . 7 + . . . + (2n - 1)(2n + 1) = \frac{n(4 n^2 + 6n - 1)}{3}\]

\[\text{ Step }  1: \]

\[P(1) = 1 . 3 = 3 = \frac{1(4 \times \left( 1 \right)^2 + 6 \times 1 - 1)}{3}\]

\[\text{ Hence, P(1) is true }  . \]

\[\text{ Step 2: }  \]

\[\text{ Let P(m) be true} . \]

\[\text{ Then,}  \]

\[1 . 3 + 3 . 5 + . . . + (2m - 1)(2m + 1) = \frac{m(4 m^2 + 6m - 1)}{3}\]

\[\text{ To prove: P(m + 1) is true}  . \]

\[\text{ That is, }  \]

\[1 . 3 + 3 . 5 + . . . + (2m + 1)(2m + 3) = \frac{(m + 1)\left[ 4(m + 1 )^2 + 6\left( m + 1 \right) - 1 \right]}{3}\]

\[ \text{ Now, P(m) is equal to: }  \]

\[1 . 3 + 3 . 5 + . . . + (2m - 1)(2m + 1) = \frac{m(4 m^2 + 6m - 1)}{3}\]

\[ \Rightarrow 1 . 3 + 3 . 5 + . . . + (2m - 1)(2m + 1) + (2m + 1)(2m + 3) = \frac{m(4 m^2 + 6m - 1)}{3} + (2m + 1)(2m + 3) \left[ \text{ Adding } (2m + 1)(2m + 3) \text{ to both sides }  \right]\]

\[ \Rightarrow P(m + 1) = \frac{m(4 m^2 + 6m - 1) + 3(4 m^2 + 8m + 3)}{3}\]

\[ \Rightarrow P(m + 1) = \frac{4 m^3 + 6 m^2 - m + 12 m^2 + 24m + 9}{3} = \frac{4 m^3 + 18 m^2 + 23m + 9}{3}\]

\[ \Rightarrow P(m + 1) = \frac{4m( m^2 + 2m + 1) + 10 m^2 + 19m + 9}{3}\]

\[ = \frac{4m(m + 1 )^2 + (10m + 9)(m + 1)}{3}\]

\[ = \frac{(m + 1)\left[ 4m(m + 1) + 10m + 9 \right]}{3}\]

\[ = \frac{(m + 1)}{3}(4 m^2 + 8m + 4 + 6m + 5)\]

\[ = \frac{(m + 1)\left[ 4(m + 1 )^2 + 6\left( m + 1 \right) - 1 \right]}{3}\]

\[\text{ Thus, P(m + 1) is true .}  \]

\[\text{ By the principle of mathematical induction, P(n) is true for all n}  \in N .\]