Find the sum of the series : 1.n+2. (n-1) + 3.(n-2)+....+(n-1).2+n.1.

Open in App

Solution

$I n + 2. (n - 1) + 3 (n - 2) + . . . . . . n .1$
$\Rightarrow t r = r (n - r)$
$∴$ $\sum_{r = 1}^{n} t r = \sum_{r = 1}^{n} r (n - r)$
$= \sum_{r = 1}^{n} (n r - r^{2})$
$= n . \frac{n (n + 1)}{2} - \frac{n (n + 1) (2 n + 1)}{6}$
$= \frac{n^{2} (n + 1)}{2} - \frac{(n^{2} + n) (2 n + 1)}{6} = \frac{n^{3} + n}{2} - (\frac{{2 n}^{3} + {3 n}^{2} + n}{6}) = \frac{n^{3} - {3 n}^{2} + 2 n}{6}$

Suggest Corrections

Similar questions

1. n + 2. (n - 1) + 3. (n - 2) + 4. (n - 3) + . . . . + (n - 1) .2 + n .1

x^{n - 1}

(1 + 2 x + 3 x^{2} + . . . . + n x^{n - 1})^{2}

1. n + 2 (n - 1) + 3 (n - 2) + . . . + n .1

n \geq 1

\frac{1}{1.2} + \frac{1}{2.3} . . . \frac{1}{n . n + 1}

\frac{{1.2}^{2} + {2.3}^{2} + {3.4}^{2} + . . . n (n + 1)^{2}}{1.2 + 2^{2} .3 + 3^{2} .4 + . . . n^{2} (n + 1)}

Find the sum of the series whose nth term is :

(i) $2 n^{3} + 3 n^{3 - 1}$

(ii) $n^{3} - 3^{n}$

(iii) $n^{(} n + 1) (n + 4)$

(iv) $(2 n - 1)^{2}$

Join BYJU'S Learning Program