Question

Please help for Q4 I’m super lost. Write an expression for the sum of four consecutive odd numbers where 2n+1 is the smallest odd number

Answer 1

Answer:

n+2

Step-by-step explanation:

This is a Typical Question on Arithmetic Progression with initial term as 2n+1

The next number in this particular series will be 2n+1+2 which is 2n+3 and so on. Thus a common difference of 2 exists.

The nth term of an Arithmetic Progression AP, bn is represented below where b1 is the first term

bn=b1 +(n-1)d where d is the common difference

For this particular series, the 4th term b4 is

b4= 2n+1+(4-1)2 0r b4 =2n+1+6=2n+7

sum of series is represented by equation: n(b1 +b4)/2 where b1 and b4 are 1st and 4th term respectively

Thus Sum required =4(2n+1+2n+7)/2=4(4n+8)/2=8n+16  0r n+2