The rth term of a series is given by

$\displaystyle

u_r=(\frac{1}{3})^{3r-2}+(\frac{1}{3})^{3r-1}

$

Express $\displaystyle \sum^{n}_{r=1}u_r$ in the form

$\displaystyle A(1-\frac{B}{27^n}) $, where A and B are constants to be found ,

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- Feb 25th 2009, 11:07 PMthereddevilsterm of a series
The rth term of a series is given by

$\displaystyle

u_r=(\frac{1}{3})^{3r-2}+(\frac{1}{3})^{3r-1}

$

Express $\displaystyle \sum^{n}_{r=1}u_r$ in the form

$\displaystyle A(1-\frac{B}{27^n}) $, where A and B are constants to be found , - Feb 26th 2009, 01:14 AMADARSH
$\displaystyle

u_r=(\frac{1}{3})^{3r-1}\frac{3}{1}+(\frac{1}{3})^{3r-1}

$

$\displaystyle

u_r=(\frac{1}{3})^{3r-1}(3+1)

$

$\displaystyle

u_r=(\frac{1}{3})^{3r-1}(4) ~= 12\frac{1}{3^{3r}} ~= \frac{12}{27^r}

$

No need to tell you that this is a GP

$\displaystyle Sum = a\frac{(r^n-1)}{r-1}$

$\displaystyle =\frac{4}{9} \frac{(1-\frac{1}{27^n})}{(1-\frac{1}{27})}

$

$\displaystyle =\frac{4 \times 27}{9\times 26} {(1-\frac{1}{27^n})}

$

I think now you can continue to (Dance)