Find the sum of $15$ terms of the series whose $n^{t h}$ term is $4 n + 1$ .

Open in App

Solution

$T_{n} = 4 n + 1$
sum $= \sum_{k = 1}^{15} (4 k + 1)$
$= \sum_{k = 1}^{15} 4 k + \sum_{k = 1}^{15} (1)$
$= 4 [1 + 2 + 3 + 4 + . . . . . . . . . . + 15]$
$= 4 \times \frac{15 \times (15 + 1)}{2} + 15$
$= 480 + 15$
$= 495$

Suggest Corrections

Similar questions

n

n t h

(2 n - 1)^{2}

Find the sum of 10 terms of the series whose $n^{t h}$ term is 3. $2^{n}$ .

15

n^{t h}

3 + 4 n

Find the sum of first 25 terms of an AP whose nth term is 1-4n.

n^{t h}

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

Join BYJU'S Learning Program