Please solve the problem for me

Find the sum of the series

3+7+14+27+52+...................up to nth terms

Results 1 to 6 of 6

- Jun 7th 2008, 02:09 AM #1

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- Jun 7th 2008, 02:30 AM #2
The nth term in the series is given by the recurrence relation $\displaystyle a_n = 2 a_{n-1} - n + 3$ with $\displaystyle a_1 = 3$. This does not define an Arithmetic Sequence or a Geometric Sequence.

So I don't know why you would imply the series is arithmetic or geometric in your descriptive title of the problem.

Once you've got the nth term of the series in terms of n, you'll have more luck finding the value of the series.

- Jun 7th 2008, 02:36 AM #3

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- Jun 7th 2008, 02:37 AM #4

- Jun 7th 2008, 02:45 AM #5
3 + 7 + 14 + 27 + 52 + ...

$\displaystyle a_n = 3\cdot 2^n + n$ (starting from $\displaystyle a_0$)

$\displaystyle S_n = \sum_{k=0}^{n}3\cdot 2^k + k$

$\displaystyle S_n = 3\sum_{k=0}^{n}2^k +\sum_{k=0}^{n}k$

$\displaystyle S_n = 3\cdot (2^{n+1}-1) + \frac{n(n+1)}{2}$

You can add +1 to $\displaystyle S_n$ if you want to sum 1+3+7+14+27+...

- Jun 7th 2008, 02:51 AM #6
The nth term in the series (the original, not the revised) is $\displaystyle \frac{3}{2} \, (2^n) + n - 1$ (starting from n = 1). Others will benefit from knowing the correct series and might have more to say. I'll just give this reply to complement my first.

And in fact, you can still use this result. Use it to get the sum and then just add 1 to the result.

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