Evaluate the following:

Question:

Evaluate the following:

(i) $\sum_{n=1}^{11}\left(2+3^{n}\right)$

(ii) $\sum_{k=1}^{n}\left(2^{k}+3^{k-1}\right)$

(iii) $\sum_{n=2}^{10} 4^{n}$

Solution:

(i) 

$S_{11}=\sum_{n=1}^{11}\left(2+3^{n}\right)$

$\Rightarrow S_{11}=\sum_{n=1}^{11} 2+\sum_{n=1}^{11} 3^{n}$

$\Rightarrow S_{11}=2 \times 11+\left(3+3^{2}+3^{3}+\ldots+3^{11}\right)$

$=22+3\left(\frac{3^{11}-1}{3-1}\right)$

$=22+\left(\frac{177147-1}{2}\right)$

$=22+265719$

$=265741$

(ii) 

$S_{n}=\sum_{k=1}^{n}\left(2^{k}+3^{k-1}\right)$

$=\sum_{k=1}^{n} 2^{k}+\sum_{k=1}^{n} 3^{k-1}$

$=\left(2+4+8+\ldots+2^{n}\right)+\left(1+3+9+\ldots+3^{n}\right)$

$=2\left(\frac{2^{n}-1}{2-1}\right)+1\left(\frac{3^{n}-1}{3-1}\right)$

$=\frac{1}{2}\left(2^{n+2}-4+3^{n}-1\right)$

$=\frac{1}{2}\left(2^{n+2}+3^{n}-5\right)$

(iii) 

$\sum_{n=2}^{10} 4^{n}=4^{2}+4^{3}+4^{4}+\ldots+4^{10}$

$=16+64+256+\ldots+4^{10}$

$=16\left(\frac{4^{9}-1}{4-1}\right)=\frac{16}{3}\left(4^{9}-1\right)$

Leave a comment

Close
faculty

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now