Question

How do i solve it And prove the answer is correct

Answer 1

we have that

Sn=24-6n

For n=1 -----> S1=24-6(1)=18

For n=2 ----> S2=24-6(2)=12

For n=3 ----> S3=24-6(3)=6

For n=4 ----> S4=24-6(4)=0

the first 4 terms are

18,12,6,0

their sum is

18+12+6+0=36

answer is option C

Problem N 2

we have

`\sum ^4_(k\mathop=2)2^k`
`\sum ^4_{k\mathop{=}2}2^k=2^2+2^3+2^4=4+8+16=28`