Let a_n be the n^ term of the series 5+8+14+23+35+50+ and S_n= _k=1^na_k.

Question

Let ${a}_{n}$ be the ${n}^{}$ term of the series $5+8+14+23+35+50+\ldots$ and ${S}_{n}=\sum _{k=1}^{n}{a}_{k}$. Then ${S}_{30}-{a}_{40}$ is equal to :

Accepted Answer

Let ${S}_{n}=5+8+14+23+\ldots .+{a}_{n}$

and ${S}_{n}=0+5+8+14+\ldots .+{a}_{n}$

On subtracting, we get

$\begin{gathered} 0=5+3+6\ldots -{a}_{n} \ \Rightarrow {a}_{n}=5+3+6+9+\ldots (n-1) \ =5+[\frac{(n-1)}{2}(6+(n-2)3)] \end{gathered}$

$\begin{gathered} =5+[\frac{(n-1)}{2}(6+3n-6)] \ =5+\frac{(n-1)(3n)}{2} \ =\frac{10+3{n}^{2}-3n}{2} \end{gathered}$

$\begin{gathered} {a}_{40}=\frac{3(40{)}^{2}-3(40)+10}{2} \ =\frac{4800-120+10}{2}=2345 \end{gathered}$

$\begin{gathered} {S}_{n}=\sum _{k=1}^{n}{a}_{k} \ \Rightarrow {S}_{30}=\frac{3\sum _{30}^{n=1}{n}^{2}-3\sum _{30}^{n=1}n+10\sum _{30}^{n=1}1}{2} \ =\frac{3	imes (30)(30+1)(60+1)}{12}-\frac{3	imes 30	imes 31}{4}+\frac{10	imes 30}{2} \end{gathered}$

$\begin{gathered} =\frac{28365-1395+300}{2}=\frac{27270}{2} \ =13635 \ 	herefore {S}_{30}-{a}_{40}=13635-2345=11290 \end{gathered}$

Solutions · 1