Question

please solve this question as soon as possible ​

Answer 1

✪ᴀɴsᴡᴇʀ✪

• $\boxed{\bf{ S_n = 2^{n+1} + \frac{3n^2}{2} + \frac{n}{2}-2}}$

☆ɢɪᴠᴇɴ☆

• nth term of a series is given by

$\:\:\:\:\:\:\:\:\:\: t_n = 2^n + 3n - 1$

★ᴛᴏ ғɪɴᴅ★

• Sum upto nth term.

⍟ᴄᴀʟᴄᴜʟᴀᴛɪᴏɴ⍟

Now,

$S_n = \sum \limits_{r =1}^{n} [2^r + 3r - 1]$

$S_n = \sum \limits_{r =1}^{n}2^r + 3\sum \limits_{r =1}^{n}r - \sum \limits_{r =1}^{n}1$

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

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

$S_n = 2^{n+1} - 2 + 3\frac{n^2}{2} + 3\frac{n}{2}- n$

$S_n = 2^{n+1} - 2 + 3\frac{n^2}{2} + \frac{n}{2}$

$S_n = 2^{n+1} + \frac{3n^2}{2} + \frac{n}{2} - 2$