Question

class 11 trignometry
plz plz solve if u know​

Answer 1

${\huge{\bf{\red{\underline{Solution:}}}}}$

${\bf{\blue{\underline{Given:}}}}$

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

${\bf{\blue{\underline{Now:}}}}$

Take L.H.S,

$\diamond \: \: {\sf{ \green{ put \: P(n )= 1}}}$

$={\sf{ [2(1) - 1][2(1) + 1]}}$

$={\sf{ [1][3]}}$

$={\sf{ 3}}$

Now take R.H.S,

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

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

$={\sf{ \frac{10 - 1}{3} }}$

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

$={\sf{ 3 }}$

L.H.S = R.HS

Result is true for n = 1

___________________________________

Assume that the result is true for n = k

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

Adding (2k+1) and (2k+3) to both side,

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

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

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

$: \implies{\sf{ \frac{k(4 {k }^{2} + 6k - 1) + 3 \times (4 {k}^{2} + 8k + 3) }{3} }}$

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

$: \implies{\sf{ \frac{4 {k}^{3} + 18 {k}^{2} - k + 24k + 9}{3} }}$

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

$: \implies{\sf{ \frac{(k + 1)(4 {k}^{2} + 14 {k} + 9)}{3} }}$

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

P (k+1) is true.

• Hence by principle of Mathematical induction p(n) is true for all n€N

Answer 2

Let ,

$\star \sf \: P(n) = 1.3 + 3.5 + 5.7 + ... + (2n - 1)(2n + 1) = \frac{n(4 {n}^{2} + 6n - 1)}{3} - - - (i)$

For n = 1 ,

LHS of eq (i) is

$\sf \hookrightarrow (2 \times 1 - 1)(2 \times 1+ 1) \\ \\ \sf \hookrightarrow 1 \times 3 \\ \\ \sf \hookrightarrow 3$

and

RHS of eq (i) is

$\sf \hookrightarrow \frac{1(4 {(1)}^{2} + 6(1)- 1)}{3} \\ \\ \sf \hookrightarrow \frac{4 + 6 - 1}{3} \\ \\ \sf \hookrightarrow \frac{9}{3} \\ \\ \sf \hookrightarrow 3$

$\sf \therefore \underline{LHS = RHS}$

Therefore , P(1) is true

Assume P(k) is true , then

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

For n = k + 1

$\sf \hookrightarrow P(k + 1) = 1.3 + 3.5 + 5.7 + ... + (2k - 1)(2k + 1) + \bigg(2(k + 1) - 1 \bigg) \bigg(2(k + 1) + 1 \bigg) = \frac{(k + 1) \bigg(4 {(k + 1)}^{2} + 6(k + 1) - 1 \bigg)}{3} \\ \\ \sf \hookrightarrow P(k + 1) = P(k) + (2k + 1 )(2 k +3) = \frac{(k + 1) \bigg( 4( {k}^{2} + 1 + 2k ) + 6k + 6 - 1 \bigg)}{3} \\ \\\sf \hookrightarrow P(k + 1) = P(k) + (2 k+ 1 )(2k +3) = \frac{(k + 1)(4 {k}^{2} + 4 + 8k + 6k + 5) }{3} \\ \\ \sf \hookrightarrow P(k + 1) = P(k) + (2k + 1)(2 k + 3) = \frac{(k + 1) (4 {k}^{2} + 14k + 9)}{3} \\ \\ \sf \hookrightarrow P(k + 1) = P(k) + (2k + 1)(2 k + 3) = \frac{4 {k}^{3} + 14 {k}^{2} + 9k + 4 {k}^{2} + 4 {k}^{2} + 14k + 9 }{3} \\ \\ \sf \hookrightarrow P(k + 1) = P(k) + (2k + 1)(2 k + 3) = \frac{4 {k}^{3} + 18 {k}^{2} + 23k + 9}{3}$

LHS :

$\sf \hookrightarrow P(k + 1) = P(k) + (2k + 1)(2 k + 3) \\ \\ \sf \hookrightarrow P(k + 1) =\frac{k(4 {k}^{2} + 6k - 1)}{3} + (2k + 1)(2k + 3) \\ \\\sf \hookrightarrow P(k + 1) = \frac{4 {k}^{2} + 6 {k}^{2} - k + 12 {k}^{2} + 18k + 6k + 9 }{3} \\ \\ \sf \hookrightarrow P(k + 1) = \frac{4 {k}^{3} + 18 {k}^{2} + 23k + 9}{3}$

$\sf \therefore \underline{LHS = RHS}$

Therefore , P(n) is true for all natural number