Maths

15. We can write the given statement as

P(n)=1²+3²+5²+ … + (2n – 1)²= $\frac{n(2n-1)(2n+1)}{3}$

forn=1

P(1)=1²=1= $\frac{1(2.1-1)(2.1+1)}{3}$

= $\frac{1(1)(3)}{3}=1$ which is true.

Consider P(k) be true for some positive integer k

P(k)=1²+3²+5²+ … + (2n – 1)²= $\frac{k(2k-1)(2k+1)}{3}$ ------------------(1)

Now, let us prove that P(k+1) is true.

Here,

1²+3²+5²+ … +(2k – 1)²+(2(k+1) –1)²

By using (1),

$\frac{k(2k-1)(2k+1)}{3}+{[2k+2-1)}^2$

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

= $\frac{(2k+1)[k(2k-1)+3(2k+1)]}{3}$

= $\frac{(2k+1)\left(2k^2-k+6k+3\right)}{3}$

= $\frac{(2k+1)\left(2k^2+5k+3\right)}{3}$

we can write as,

= $\frac{(2k+1)\left(2k^2+2k+3k+3\right)}{3}$

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

= $\frac{(2k+1)(2k+3)(k+1)}{3}$

= $\frac{(2k+1)(k+1)(2k+3)}{3}$

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

P(k+1) is true whenever P(k) is true.

Hence, from the principle of mathematical induction, the P(n) is true for all natural number n.

93. Kindly go through the solution

14. Let the given statement be P(n) i.e.,

P(n)= $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ … $\left(1+\frac{1}{n}\right)=(n+1)$

If n =1

P(1)= $\left(1+\frac{1}{1}\right)$ = 2 =1+1= 2

which is true.

Assume that P(k) is true for some positive integer k i.e.,

P(k): $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ … $\left(1+\frac{1}{k}\right)=(k+1)$ .---------------------(1)

Now, let us prove that P(k+1) is true.

Here,

P(k+1)= $\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)\left(1+\frac{1}{3}\right)$ … $\left(1+\frac{1}{k}\right)+\left(1+\frac{1}{(k+1)}\right)$

By using (1), we get

(k+1). $\left(1+\frac{1}{k+1}\right)$

L.C.M.=(k+1). $\left(\frac{k+1+1}{k+1}\right)$

= (k+1)+1

? P(k+1) is true whenever P(k) is true.

Therefore from the principle of mathematical induction the P(n) is true for all natural numbers n.

In a particular solution, there are no arbitrary constant.

Hence, option (D) is correct.

92. Kindly go through the solution

The number of arbitrary constant is general solution of D.E of 4^th order is four.

$\therefore$ Option (D) is correct.

13. We can write given statement as

P(n): $\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2n+1)}{n^2}\right)={(n+1)}^2$

If n=1, we get

P(1): $\left(1+\frac{3}{1}\right)$ =4=(1+ 1)²=2²=4

which is true.

Consider P(k) be true for some positive integer k.

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2k+1)}{k^2}\right)={(k+1)}^2$ (1)

Now, let us prove that P(k+1) is true.

$\left(1+\frac{3}{1}\right)\left(1+\frac{5}{4}\right)\left(1+\frac{7}{9}\right)$ … $\left(1+\frac{(2k+1)}{k^2}\right)+\left(1+\frac{(2(k+1)+1)}{{(k+1)}^2}\right)$

By using (1)

=(k+1)²$\left(1+\frac{2(k+1)+1}{{(k+1)}^2}\right)$

=(k+1)² $\left[\frac{{(k+1)}^2+2(k+1)+1}{{(k+1)}^2}\right]$

=(k+1)²+2(k+1)+1

={(k+1)+1}²

P(k+1) is true whenever P(k) is true.

Therefore, by principle of mathematical induction, the P(n) is true for all natural number n.

Kindly go through the solution

91. Given, $x=a\left(cost+logtan\frac{t}{2}\right)y=asint$

Differentiating w r t we get,

$\frac{dx}{dt}=a\frac{d}{dt}\left[cost+log\left(tan\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{1}{tan\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\frac{d}{dt}\left(tan\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{1}{tan\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}.se{c}^2\frac{t}{2}\frac{d}{dt}\left(\frac{t}{2}\right)\right]$

$=a\left[-sint+\frac{cos\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}{sin\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}*\frac{1}{co{s}^2\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}*\frac{1}{2}\right]$

$=a\left[-sint+\frac{1}{2sin\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.cos\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=a\left[-sint+\frac{1}{sin2*\raisebox{1ex}{$t$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\right]$

$=a\left[-sint+\frac{1}{sint}\right]=a\left[\frac{1-si{n}^{2}t}{sint}\right]$

$=a\frac{co{s}^{2}t}{sint}\left\{?1=co{s}^{2}x+si{n}^{2}x\right\}$

$\frac{\mathrm{bd}y}{dt}=\frac{d}{dt}\left(asint\right)=acost$

$\therefore \frac{dy}{dx}=\frac{\raisebox{1ex}{$dy$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}{\raisebox{1ex}{$dx$}\!\left/ \!\raisebox{-1ex}{$dt$}\right.}=\frac{acost}{\raisebox{1ex}{$aco{s}^{2}t$}\!\left/ \!\raisebox{-1ex}{$sint$}\right.}=\frac{sint}{cost}=tant$

12. Let the given statement be P(n) i.e.,

P(n)=a+ar+ar²+ … +ar^n-1== $\frac{a\left(r^n-1\right)}{r-1}$

If n = 1, we get

P(1)=a= $\frac{a\left(r^1-1\right)}{r-1}$ =a

which is true.

Consider P(k) be true for some positive integer k

a+ar+ar²+ … +ar^k-1= $\frac{a\left(r^k-1\right)}{r-1}$ (1)

Now, let us prove that P(k+1) is true.

Here, {a+ar+ar²+ … +ar^k-1}+ar^{(k+1) –1}

By using (1),

= $\frac{a\left(r^k-1\right)}{r-1}+a{r}^k$

= $\frac{a\left(r^k-1\right)+a{r}^k(r-1)}{r-1}$

= $\frac{a{r}^k-a+a{r}^{k+1}-a{r}^k}{r-1}$

= $\frac{a{r}^{k+1}-a}{r-1}$

= $\frac{a\left(r^{k+1}-1\right)}{r-1}$

P(k+1) is true whenever P(k) is true.

Therefore, by the principle of mathematical induction, statement P(n) is true for all natural numbers i.e.,