(d) : $\left(S1\right) : \underset{n\to \infty }{\lim }\frac{1}{n^2}(2+4+6+...+2n)$

$=\underset{n\to \infty }{\lim }\frac{1}{n^2}\times 2(1+2+3+...+n)$

$=\underset{n\to \infty }{\lim }\frac{1}{n^2}\times 2\times \frac{n(n+1)}{2}=\underset{n\to \infty }{\lim }(1+\frac{1}{n})=1$

$\therefore \left(S1\right) is true.$

$\left(S2\right) : \underset{n\to \infty }{\lim }\frac{1}{n^{16}}(1^{15}+2^{15}+3^{15}+....+n^{15})=\underset{n\to \infty }{\lim }\frac{1}{n^{16}}\sum _{r=1}^n{r}^{15}$

$=\underset{n\to \infty }{\lim }\frac{1}{n}\sum _{r=1}^n{\left(\frac{r}{n}\right)}^{15}={\int }_0^1{x}^{15}dx=|\frac{x^{16}}{16}{|}_0^1=\frac{1}{16}$

(S2) is also true.

(c) : $Let L = \underset{n\to \infty }{\lim }\left\{\right(2^{\frac{1}{2}}-2^{\frac{1}{3}}\left)\right(2^{\frac{1}{2}}-2^{\frac{1}{5}})....(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}}\left)\right\}$

$By Sandwich Theorem,$

${(2^{\frac{1}{2}}-2^{\frac{1}{3}})}^n<(2^{\frac{1}{2}}-2^{\frac{1}{3}})(2^{\frac{1}{2}}-2^{\frac{1}{5}})(2^{\frac{1}{2}}-2^{\frac{1}{7}})....(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}})<{(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}})}^n$

$\Rightarrow \underset{n\to \infty }{\lim } {(2^{\frac{1}{2}}-2^{\frac{1}{3}})}^n<L<\underset{n\to \infty }{\lim } {(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}})}^n$

$As, \underset{n\to \infty }{\lim } {(2^{\frac{1}{2}}-2^{\frac{1}{3}})}^n=0 and \underset{n\to \infty }{\lim } {(2^{\frac{1}{2}}-2^{\frac{1}{2n+1}})}^n=0$

$\therefore L = 0$

(d) : $\underset{x\to 0}{\lim }\left[\right(\frac{(1-{\cos }^{2}3x)}{{\cos }^3\left(4x\right)}\left)\right(\frac{{\sin }^3\left(4x\right)}{({\log }_e{(2x+1))}^5}\left)\right]$

$=\underset{x\to 0}{\lim }[\frac{{\sin }^2\left(3x\right)}{{\cos }^{3}4x}\times \frac{{\sin }^3\left(4x\right)}{[{\log }_e{(2x+1)]}^5}]$

$=\underset{x\to 0}{\lim }\left[\frac{{\displaystyle \frac{{\sin }^2\left(3x\right)}{{\left(3x\right)}^2}}\times {\displaystyle \frac{{\sin }^3\left(4x\right)}{{\left(4x\right)}^3}}\times {\left(3x\right)}^2\times {\left(4x\right)}^3}{{\cos }^3\left(4x\right)·{\left[{\displaystyle \frac{{\log }_e(2x+1)}{2x}}\right]}^5\times {\left(2x\right)}^5}\right]$

$=\frac{9\times 64}{32}=18$

(d) : $Given, a{x}^2+bx+1 = 0 has roots \alpha , \beta , then x^2+bx+a=0 has roots \frac{1}{\alpha },\frac{1}{\beta }.$

$Now, \underset{x\to \frac{1}{\alpha }}{\lim } {\left(\frac{1-\cos (x^2+bx+a)}{2{(1-\alpha x)}^2}\right)}^{1/2}$

$=\underset{x\to \frac{1}{\alpha }}{\lim } {\left(\frac{2{\sin }^2\left({\displaystyle \frac{x^2+bx+a}{2}}\right)}{2{(1-\alpha x)}^2}\right)}^{1/2}$

$=\underset{x\to \frac{1}{\alpha }}{\lim } {\left(\frac{2{\sin }^2\left({\displaystyle \frac{(x-{\displaystyle \frac{1}{\alpha }})(x-{\displaystyle \frac{1}{\beta }})}{2}}\right)}{2{(1-\alpha x)}^2}\right)}^{1/2}$

$=\underset{x\to \frac{1}{\alpha }}{\lim } {\left(\frac{{\displaystyle \frac{{\sin }^2\left(\frac{(1-\alpha x)(1-\beta x)}{2\alpha \beta }\right)}{{\left(\frac{(1-\alpha x)(1-\beta x)}{2\alpha \beta }\right)}^2}}\times {\left({\displaystyle \frac{(1-\alpha x)(1-\beta x)}{2\alpha \beta }}\right)}^2}{{(1-\alpha x)}^2}\right)}^{1/2}$

$=\underset{x\to \frac{1}{\alpha }}{\lim } \left(\frac{1-\beta x}{2\alpha \beta }\right)=\left(\frac{\alpha -\beta }{2{\alpha }^2\beta }\right)=\frac{1}{2\alpha }(\frac{1}{\beta }-\frac{1}{\alpha })$

$=\frac{1}{k}(\frac{1}{\beta }-\frac{1}{\alpha }) (\Rightarrow Given)$

$So, k = 2\alpha$

(c) : $\underset{x\to 0}{\lim }\frac{a{e}^{ax}+b\sin bx-{\displaystyle \frac{c}{2}}{e}^{-cx}+{\displaystyle \frac{c^{2}x}{2}}{e}^{-cx}}{2\sin 2x}=17$

$\Rightarrow a-\frac{c}{2}=0 \therefore a=\frac{c}{2}$

Again, Applying (L'Hospital rule)

$\underset{x\to 0}{\lim }\frac{a^2{e}^{ax}+b^2\cos bx+{\displaystyle \frac{c^2}{2}}{e}^{-cx}-{\displaystyle \frac{c^{3}x}{2}}{e}^{-cx}+{\displaystyle \frac{c^2}{2}}{e}^{-cx}}{4\cos 2x}=17$

$\Rightarrow \frac{a^2+b^2+{\displaystyle \frac{c^2}{2}}+{\displaystyle \frac{c^2}{2}}}{4}=17$

$\Rightarrow a^2+b^2+c^2=68\Rightarrow a^2+b^2+4a^2=68$

$\Rightarrow 5a^2+b^2=68$