## Tuesday, 22 September 2009

### Powers of three.... Continue the Pattern

1 --------------------------- = $3^{0}$
1 + 2------------------------ =$3^{1}$
2 + 3 + 4 ----------------- =$3^{2}$
2 + 3 + 4 + 5 + 6 + 7 ------- =$3^{3}$
5 + 6 + 7 + .... + 13 ------- =$3^{4}$
5 + 6 + 7 + .... + 22 ------- =$3^{5}$
$\sum_{n=14}^{40}{n} =3^{6}$
$\sum_{n=14}^{67}{n} =3^{7}$
$\sum_{n=41}^{121}{n} =3^{8}$

and then???? Can you write a general expression?

From the Mathematical Spectrum, 1983, submitted by L. B. Dutta of Keshabpur, Bangladesh

#### 1 comment:

I. J. Kennedy said...

Start at (3^(n div 2) + 1) / 2, add 3^(n div 2) (n even) or 2*3^(n div 2) (n odd) consecutive integers, and you get 3^n.