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sergij07 [2.7K]
3 years ago
13

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 od

d number

Mathematics
1 answer:
sveta [45]3 years ago
6 0

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

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General Formulas and Concepts:

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  D. Both functions are decreasing at the same average rate on that interval

Step-by-step explanation:

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Function f decreases by 60 units from f(0) = 64 to f(4) = 4 on the interval x = [0, 4]. Function g decreases by 60 units from g(0) = 75 to g(4) = 15 on the same interval. The average rate of change is the amount of decrease divided by the interval width. Those values are the same for both functions.

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