Progression

**MichaelLight** · March 2nd 2011, 07:57 PM

Need help with number (9)..

Thanks...

**emakarov** · March 2nd 2011, 11:40 PM

If $S_n=3n^2+4n$ , then $V_n=S_n-S_{n-1}$ .

**Soroban** · March 3rd 2011, 05:28 AM

Hello, MichaelLight!

$\displaystyle \text{9. If }\sum^n_{r=1}V_r \,=\,3n^2 + 4n,\,\text{find the }n^{th}\text{ term of the series.}$

$\text{Hence, show that the series is an arithmetic series.}$

Since the sum of the terms is a quadratic, the sequence must be linear.

The general term would be: / $V_r \:=\:ar + b$

Then: . $\displaystyle \sum^n_{r=1}V_r \;=\;\sum^n_{r=1}(ar + b) \;=\;a\sum^n_{r=1}r + b\sum^n_{r=1}1$

. . . $=\;a\frac{n(n+1)}{2} + bn \;=\;\frac{a}{2}n^2 + \frac{a+2b}{2}n \;\;\text{ which equals }3n^2 + 4n$

So we have: . $\begin{Bmatrix} \frac{a}{2} \;=\;3 \\ \\[-3mm] \frac{a+2b}{2} \;=\;4\end{Bmatrix} \quad\Rightarrow\quad a =6,\;b=1$

. . Therefore: . $V_r \:=\:6r + 1$

The sequence is: . $7, 13, 19, 25, 31\;\hdots$ . an arithmetic series

. . with first term $a = 7$ and common difference $d = 6.$

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