Question

Find the sum of the following sequence:

0(m-1) + 1(m-2) + 2(m-3) + 3(m-4) +++++ (m-3)2 + (m-2)1 + (m-1)0.

or

Answer 1

Answer:

The answer is $\frac{m(m-4)(m+1)}{6}$

Step-by-step explanation:

General term of series is

$a_r=r[m-(r+1)], r=0,1,2,...,m+1$

    $=(m-1)r-r^2$

Thus the required sum

     $=\displaystyle \sum_{r=1}^{m}a_r=\displaystyle \sum_{r=1}^{m}[(m-1)r-r^2]\\=(m-1)\displaystyle \sum_{r=1}^{m}r-\displaystyle \sum_{r=1}^{m} r^2\\=(m-1)\times \frac{(m)(m+1)}{2}-\frac{m(m+1)(2m+1)}{6}$

    (  since value of first and last terms are anyway zero )

     $= \frac{(m-1)(m)(m+1)}{2}-\frac{(m)(m+1)(2m+1)}{6}\\=m(m+1)(\frac{m-1}{2}-\frac{2m+1}{6})\\=m(m+1)(\frac{3(m-1)-(2m+1)}{6})\\=m(m+1)(\frac{3m-3-2m-1}{6})\\=m(m+1)(\frac{m-4}{6})=\frac{m(m-4)(m+1)}{6}$