Looks tough

Let $f\left( x \right) ={(3x^2-2x+1)}^{1023}={n}_{0}x^{2046}+{n}_{1}x^{2045}+\dots + {n}_{2046}$ for unknown coefficients of the form ${n}_{i}$ . We are trying to find the value of $\displaystyle \sum _{ i=0 }^{ 2046 }{ { n }_{ i } }$ , and this sum can be evaluated by setting $x=1$ .

$\sum _{ i=0 }^{ 2046 }{ { n }_{ i } } = f\left( 1 \right) = {(3(1)^2-2(1)+1)}^{1023} = \large \color{#20A900}{\boxed{2^{1023}}}$

This seems like an easy problem in difficult problem's clothes. The coefficients are $3$ , $-2$ and $1$ . Adding them up gives us $2$ . We just have to maintain the exponent, thus, $\boxed{2^{1023}}$

$3x^{2}- 2x+1^{1023}$ To find the sum of coefficients, just put the value of x=1 (3(1)^2 -2(1) +1)^1023 => (3-2+1)^1023 = 2^1023

it is very easy problem