JEE Main 2024 — 30 January, Shift 1 — Question 30

$\alpha ={{1}^{2}}+{{4}^{2}}+{{8}^{2}}\ldots$

${{\text{t}}_{\text{n}}}=\text{a}{{\text{n}}^{2}}+\text{bn}+\text{c}$

$1=\text{a}+\text{b}+\text{c}$

$4=4a+2b+c$

$8=9a+3b+c$

On solving we get, $\text{a}=\frac{1}{2},\text{ b}=\frac{3}{2},\text{c}=-1$ $\alpha =\sum _{\text{n}=1}^{10}{{\left( \frac{{{\text{n}}^{2}}}{2}+\frac{3\text{n}}{2}-1 \right)}^{2}}$

$4\alpha =\sum _{n=1}^{10}{{\left( {{n}^{2}}+3n-2 \right)}^{2}},\beta =\sum _{n=1}^{10}{{n}^{4}}$

$4\alpha -\beta =\sum _{n=1}^{10}\left( 6{{n}^{3}}+5{{n}^{2}}-12n+4 \right)=55\left( 353 \right)+40$