Question

$1.3 + 3.5 + 5.7 + ... + (2n - 1)(2n + 1) = \frac{n(4 {n}^{2} + 6n - 1) }{3} . \: Prove \: by \: Principle \: of \: Mathematical \: Induction .$
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Answer 1

$\large\underline{\sf{Solution-}}$

Given to Prove

$\rm :\longmapsto\:1.3 + 3.5 + 5.7 + ... + (2n - 1)(2n + 1) = \frac{n(4 {n}^{2} + 6n - 1) }{3}$

Let assume that,

$\rm P(n): \:1.3 + 3.5 + 5.7 + ... + (2n - 1)(2n + 1) = \dfrac{n(4 {n}^{2} + 6n - 1) }{3}$

Step :- 1 For n = 1

$\rm P(1): \:1.3 = \dfrac{1(4 {(1)}^{2} + 6 - 1) }{3}$

$\rm P(1): \:3 = \dfrac{(4 + 6 - 1) }{3}$

$\rm P(1): \:3 = \dfrac{(10- 1) }{3}$

$\rm P(1): \:3 = \dfrac{9 }{3}$

$\rm P(1): \:3 = 3$

$\bf\implies \:P(n) \: is \: true \: for \: n = 1$

Step :- 2 Let assume that P(n) is true for n = k, where k is natural number.

$\rm P(n): \:1.3 + 3.5 + 5.7 + ... + (2k - 1)(2k + 1) = \dfrac{k(4 {k}^{2} + 6k - 1) }{3}$

Step :- 3 We have to prove that P(n) is true for n = k + 1

$\rm P(k + 1): \:1.3 + 3.5 + 5.7 + ... + (2k + 1)(2k + 3) = \dfrac{(k + 1)(4 {(k + 1)}^{2} + 6(k + 1) - 1) }{3}$

Consider, LHS

$\rm P(k + 1): \:1.3 + 3.5 + 5.7 + ... + (2k + 1)(2k + 3)$

On substituting the value from Step 2, we get

$\rm \:  =  \:  \: \dfrac{k(4 {k}^{2} + 6k - 1) }{3} + (2k + 1)(2k + 3)$

$\rm \:  =  \:  \: \dfrac{k(4 {k}^{2} + 6k - 1) + 3 (2k + 1)(2k + 3)}{3}$

$\rm \:  =  \:  \: \dfrac{4 {k}^{3} + 6 {k}^{2} - k + 3( {4k}^{2} + 6k + 2k + 3)}{3}$

$\rm \:  =  \:  \: \dfrac{4 {k}^{3} + 6 {k}^{2} - k + 3( {4k}^{2} + 8k + 3)}{3}$

$\rm \:  =  \:  \: \dfrac{4 {k}^{3} + 6 {k}^{2} - k + {12k}^{2} + 24k + 9}{3}$

$\rm \:  =  \:  \: \dfrac{4 {k}^{3} +18{k}^{2} + 23k + 9}{3}$

can be rewritten as

$\rm \:  =  \:  \: \dfrac{ {4k}^{3} + {14k}^{2} + {4k}^{2} + 9k + 14k + 9}{3}$

$\rm \:  =  \:  \: \dfrac{ {4k}^{3} + {14k}^{2} + 9k + {4k}^{2} + 14k + 9}{3}$

$\rm \:  =  \:  \: \dfrac{k( {4k}^{2} + 14k + 9) + 1( {4k}^{2} + 14k + 9)}{3}$

$\rm \:  =  \:  \: \dfrac{( {4k}^{2} + 14k + 9) (k+ 1)}{3}$

can be rewritten as

$\rm \:  =  \:  \: \dfrac{( {4k}^{2} + 8k + 6k+ 10 - 1) (k+ 1)}{3}$

$\rm \:  =  \:  \: \dfrac{( {4k}^{2} + 8k + 6k+ 6 + 4 - 1) (k+ 1)}{3}$

$\rm \:  =  \:  \: \dfrac{( {4k}^{2} + 8k + 4+ 6k + 6 - 1) (k+ 1)}{3}$

$\rm \:  =  \:  \: \dfrac{( 4({k}^{2} + 2k + 1)+ 6(k + 1) - 1) (k+ 1)}{3}$

$\bf\implies \:P(n) \: is \: true \: for \: n = k + 1$

Hence,

By the Process of Principal of Mathematical Induction,

$\rm :\longmapsto\:1.3 + 3.5 + 5.7 + ... + (2n - 1)(2n + 1) = \frac{n(4 {n}^{2} + 6n - 1) }{3}$

Answer 2

Answer:

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