Mathematics - Sequence & Series Question with Solution | TestHub

$\left(1^2+2^2+\dots +n^2\right)-\left(a_1+a_2+\dots +a_n\right)=\frac{1}{3}n\left(n^2-1\right)$

Replacing n by ( $\mathrm{n}-1$ ), then

$$\left(1^2+2^2+\dots +(\mathrm{n}-1{)}^2\right)-\left({\mathrm{a}}_1+{\mathrm{a}}_2+\dots +{\mathrm{a}}_{\mathrm{n}-1}\right)=\frac{1}{3}(\mathrm{n}-1)\left((\mathrm{n}-1{)}^2-1\right)$$

Subtracting (ii) from (i)

${\mathrm{n}}^2-{\mathrm{a}}_{\mathrm{n}}={\mathrm{n}}^2-\mathrm{n}$ $\Rightarrow {\mathrm{a}}_{\mathrm{n}}=\mathrm{n}\Rightarrow {\mathrm{a}}_7=7$ Ans.