Sequences and Series

Given, $a_0=1, a_{n+1}=3 n^2+n+a_n$ .... (i)

From Eq.(i), we get

$\begin{array}{ll} & a_1=3(0)^2+0+a_0=1 \\ \Rightarrow \quad & a_1=1 \\ & a_2=3(1)^2+1+a_1=3+1+1=5 \\ \Rightarrow \quad & a_2=5 \end{array}$

$\begin{aligned} & a_3=3(2)^2+2+a_2=12+2+5=19 \\ & \Rightarrow \quad a_3=19 \\ \end{aligned}$

Now, check the option for $n=3$ which expression satisfy by $a_3=19$

Option (a), $a_n=n^3+n^2+1$

Put $n=3$, then

$\begin{array}{rlrl} a_3 & =(3)^3+(3)^2+1 \\ & =27+9+1=37 \\ \because \quad a_3 & \neq 37 \\ \therefore \quad a_n & \neq n^3+n^2+1 \\ & \text { Option (b), } a_n =n^3-n^2+1 \end{array}$

Option (b), $a_n=n^3-n^2+1$

Put $n=3$, then

$\begin{aligned} a_3 & =(3)^3-(3)^2+1 \\ & =27-9+1=19 \\ \therefore \quad &a_n =n^3-n^2+1 \end{aligned}$

$\begin{aligned} & \text { Given, } x^2-\sqrt{3} x+1=0 \\ & \Rightarrow \quad x=\frac{\sqrt{3} \pm \sqrt{3-4}}{2} \\ & x=\frac{\sqrt{3}}{2} \pm \frac{1}{2} i=\cos \frac{\pi}{6} \pm i \sin \frac{\pi}{6} \\ & \Rightarrow \quad x^{2 n}=\cos \frac{2 n \pi}{6} \pm i \sin \frac{2 n \pi}{6} \\ & \Rightarrow \quad x^{2 n}=\cos \frac{n \pi}{3} \pm i \sin \frac{n \pi}{3} \\ & \text { and } x^{-2 n}=\cos \frac{n \pi}{3} \pm i \sin \frac{n \pi}{3} \\ & \operatorname{Now},\left(x^n-\frac{1}{x^n}\right)^2=\left(x^{2 n}+\frac{1}{x^{2 n}}-2\right) \\ & =-2+2 \cos \frac{n \pi}{3} \\ & \Rightarrow \sum_\limits{n=1}^{24}\left(x^n-\frac{1}{x^n}\right)^2=\sum_\limits{n=1}^{24}\left(-2+2 \cos \frac{n \pi}{3}\right) \\ & =-48+2\left[\cos \frac{\pi}{3}+\cos \frac{2 \pi}{3}+\ldots+\cos \frac{24 \pi}{3}\right] \\ & =-48+2\left[\frac{\cos \left(\frac{2 \pi}{3}+\frac{23 \pi}{6}\right) \cdot \sin \frac{24 \pi}{6}}{\sin \frac{\pi}{6}}\right] \\ & =-48+2 \cdot 0=-48 \\ & \end{aligned}$

$\because$ $p,q,r,s$ are in AP.

Let $p = a,q = a + d,r = a + 2d,s = a + 3d$

Given, $p,q$ are roots of

${x^2} - 2x + A = 0$

Then, $p + q = 2$ and $pq = A$

$ \Rightarrow 2a + d = 2$ .... (i)

Also, r and s are roots of ${x^2} - 18x + B = 0$

Then, $r + s = 18$ and $rs = B$

$ \Rightarrow 2a + 5d = 18$ ..... (ii)

Solving Eqs. (i) and (ii), we get

$a = - 1,d = 4$

Now, $pq = A$ (product of roots)

or $A = a(a + d) = ( - 1)( - 1 + 4)$

$A = - 3$

and $rs = B$

or $B = (a + 2d)(a + 3d)$

$ = ( - 1 + 8)( - 1 + 12) = 77$

$\therefore$ $A = - 3,B = 77$

$f(x)=x^3+a x^2+b x+c$

Let roots of $f(x)$ are $\alpha, \beta, \gamma$ and since roots are in AP.

$\therefore \quad 2 \beta=\alpha+\gamma$

Now, Sum of roots

$\begin{aligned} \alpha+\beta+\gamma & =-a \\ \text { or }2 \beta+\beta & =-a \text { or } \beta=-\frac{a}{3} \end{aligned}$

It is given that, roots are integers.

$\therefore \beta$ is an integer when $a$ is multiple of 3 .

Option (a), (c) and (d) are multiple of 3.

$\Rightarrow \quad a \neq 1214$

Option (b) is correct.

Given, a + 2b + 3c = 12 $\forall$ a, b, c $\in$ R⁺

As, AM $\ge$ GM

${{a + b + b + c + c + c} \over 6} \ge \root 6 \of {a{b^2}{c^3}} $

$ \Rightarrow {{12} \over 6} \ge \root 6 \of {a{b^2}{c^3}} \Rightarrow a{b^2}{c^3} \le {2^6}$

Hence, the maximum value of ab²c³ is 2⁶.

a_n of the series $ = n{(2n + 1)^2}\forall n \in E$

${a_n} = n[4{n^2} + 4n + 1] \Rightarrow {a_n} = 4{n^3} + 4{n^2} + n$

${S_n} = \sum {{a_n} = 4\sum {{n^3} + 4\sum {{n^2} + \sum n } } } $

$ = 4{\left( {{{n(n + 1)} \over 2}} \right)^2} + {{4n(n + 1)(2n + 1)} \over 6} + {{n(n + 1)} \over 2}$

$ = {{n(n + 1)} \over 2}\left[ {2{n^2} + 2n + {{4(2n + 1)} \over 3} + 1} \right]$

$ = {{n(n + 1)} \over 2}\left[ {{{6{n^2} + 6n + (8n + 4) + 3} \over 3}} \right]$

$ = {{n(n + 1)} \over 2}\left[ {{{6{n^2} + 14n + 7} \over 3}} \right] = {n \over 6}(n + 1)(6{n^2} + 14n + 7)$