Given expression is $\left\{(a_{1})^{1}\times(a_{2})^{2}\times...\times(a_{20})^{20}\right\}$<$\left\{a_{21}\times a_{22}\times...\times a_{20+m}\right\}$,
$\left\{(a_{1})^{1}\times(a_{2})^{2}\times...\times(a_{20})^{20}\right\}$ = $\left\{3^1\times3^4\times3^9...\times3^{400}\right\}$
Sum of square of n natural numbers is $\frac{n\cdot\left(n+1\right)\cdot\left(2n+1\right)}{6}$
= $3^{\dfrac{\left(20\cdot21\cdot41\right)}{6}}$ = $3^{2870}$
On right hand side of inequlaity we have $\left\{a_{21}\times a_{22}\times...\times a_{20+m}\right\}$
= $3^{21}\times3^{22}\times...\times3^{20+m}$ = $3^{21+22+...+20+m}$
Using the sum of the first (n) natural numbers,
$1+2+\cdots+n = \frac{n(n+1)}{2}$
$21 + 22 + \cdots + (20+m)$
= $1+2+\cdots+(20+m) - (1+2+\cdots+20)$
$1+2+\cdots+(20+m)=\frac{(20+m)(21+m)}{2}$
$1+2+\cdots+20 = \frac{20\cdot21}{2} = 210$
So, $21+22+\cdots+(20+m)$
= $\frac{(20+m)(21+m)}{2} - 210$
Expanding, $(20+m)(21+m)=m^2+41m+420$
Thus, $\frac{m^2+41m+420}{2}-210$
$= \frac{m^2+41m}{2}$
Since the bases are equal, we must compare the powers.
$2870$<$\frac{m^2+41m}{2} \Rightarrow 5740$
Here, we can put in the option to check the minimum value that satisfies the inequality.
56: We get 5740<5264. This is false
57: We get 5740<5586. This is false
58: We get 5740<5742. This is the minimum possible value.
