Question

If ${\left(16\right)}^{15}+2\cdot 17{\left(16\right)}^{14}+3{\left(17\right)}^2{\left(16\right)}^{13}+....+16{\left(17\right)}^{15}=m{\left(16\right)}^{15}$ then value of $\sqrt{m+33}$ equals

• A. $15$
• B. $16$
• C. $17$
• D. $18$

Answer 1

Answer: (C)

$\because m{\left(16\right)}^{15}={\left(16\right)}^{15}+2\cdot 17{\left(16\right)}^{14}+3{\left(17\right)}^2{\left(16\right)}^{13}+....+16^1{\left(17\right)}^{15}$
$\therefore m=1+\frac{2\cdot 17{\left(16\right)}^{14}}{{\left(16\right)}^{15}}+3\cdot \frac{{\left(17\right)}^2{\left(16\right)}^{13}}{{\left(16\right)}^{15}}+...+\frac{16{\left(17\right)}^{15}}{{\left(16\right)}^{15}}$
or $m=1+2\left(\frac{17}{16}\right)+3{\left(\frac{17}{16}\right)}^2+...+16{\left(\frac{17}{16}\right)}^{15}$
or $m=1+2x+3x^2+...+16x^{15}$ (where $x=\frac{17}{16}$)
$=\frac{d}{dx}\left(x^1+x^2+....+x^{16}\right)$
$=\frac{d}{dx}\left[\frac{x\left(x^{16}-1\right)}{x-1}\right]$
$=\frac{\left(x-1\right)\left(17x^{16}-1\right)-x\left(x^{16}-1\right)}{{\left(x-1\right)}^2}=\frac{16x^{17}-17x^{16}+1}{{\left(x-1\right)}^2}$
$\Rightarrow m=\frac{1}{{\left(x-1\right)}^2}$ ($\because 16x^{17}=17x^{16}$ where $x=\frac{17}{16}$)
$\Rightarrow m=256$
$\therefore m+33=289\Rightarrow \sqrt{m+33}=17$