Question

$\huge\bold\red{\sum\limits_{n=1}^8 2 {}^{n - 1}}$


❖ᴏɴʟʏ ᴘʀᴏᴘᴇʀ ꜱᴏʟᴠᴇᴅ ᴀɴꜱᴡᴇʀ ᴡɪᴛʜ ɢᴏᴏᴅ ᴇxᴘʟᴀɴᴀɪᴏɴ ɴᴇᴇᴅᴇᴅ
❖ ɴᴏ ꜱᴘᴀᴍᴍɪɴɢ
❖ᴏɴʟʏ ꜰᴏʀ ᴍᴏᴅᴇʀᴀᴛᴏʀꜱ, ʙʀᴀɪɴʟʏ ꜱᴛᴀʀꜱ ᴀɴᴅ ᴏᴛʜᴇʀ ʙᴇꜱᴛ ᴜꜱᴇʀꜱ​​​​

Answer 1

$\bold \red{\sum_{n=1}^{8} 2^{n - 1}}$

$\small \bold{\color{red}{\left(\sum_{n=1}^{8} 2^{n - 1}\right)}=\color{red}{\left(1 + 2 + 4 + 8 + 16 + 32 + 64 + 128\right)}}$

$\color{red}{\left(\sum_{n=1}^{8} 2^{n - 1}\right)}=\color{red}{\left(255\right)}$

Hence,

$\bold \red{\sum_{n=1}^{8} 2^{n - 1}=255}$