Let _k=1^na_k= n^2+ n.

Question

Let $\sum _{k=1}^{n}{a}_{k}=\alpha {n}^{2}+\beta n$. If ${a}_{10}=59$ and ${a}_{6}=7{a}_{1}$, then $\alpha +\beta $ is equal to :

Accepted Answer

Let ${S}_{n}=\sum _{k=1}^{n}{a}_{k}=\alpha {n}^{2}+\beta n$

${a}_{1}+{a}_{2}+{a}_{3}+{a}_{4}+\ldots {a}_{n}=\alpha {n}^{2}+\beta n$

Since the sum ${S}_{n}$ is a quadratic expression in $n$ with no constant term then the sequence ${a}_{k}$ is in A.P (arithmetic progression).

Then $n$th term of the sequence can be found using the formula

${a}_{n}={S}_{n}-{S}_{n-1}$

$\begin{gathered} {a}_{n}=\alpha {n}^{2}+\beta n-[\alpha (n-1{)}^{2}+\beta (n-1)] \ =\alpha {n}^{2}+\beta n-[\alpha ({n}^{2}-2n+1)+\beta n-\beta ] \ =\alpha {n}^{2}+\beta n-\alpha {n}^{2}+2\alpha n-\alpha -\beta n+\beta \end{gathered}$

$\Rightarrow$ ${a}_{n}=\beta -\alpha +2\alpha n-(1)$

It is given that ${a}_{10}=59$, substitute $n=10$ in equation (1).

${a}_{10}=\beta -\alpha +2\alpha (10)=59$

$\Rightarrow$ $19\alpha +\beta =59-(2)$

also, ${a}_{6}=7{a}_{1}$ (given)

substitute $n=6$ and $n=1$ in equation (1).

$\beta -\alpha +2\alpha (6)=7(\beta -\alpha +2\alpha (1))$

$\Rightarrow$ $11\alpha +\beta =7\alpha +7\beta $

$\Rightarrow$ $4\alpha =6\beta $

$\Rightarrow$ $2\alpha =3\beta \Rightarrow \alpha =\frac{3\beta }{2}$

Substitute $\alpha =\frac{3\beta }{2}$ in equation (2)

$19(\frac{3\beta }{2})+\beta =59$

$\Rightarrow$ $57\beta +2\beta =118$

$\Rightarrow$ $59\beta =118$

$\Rightarrow$$\beta =\frac{118}{59}=2$

And $\alpha =\frac{3\beta }{2}=\frac{3	imes 2}{2}=3$

So, value of $\alpha +\beta =3+2=5$.

Solutions · 1