Answer:
Add the 2 numbers up together. Thatś how I have learned to do it anyway. maybe someone else has something different.
Step-by-step explanation:
It is true that the product of two consecutive even integers are always one less than the square of their average.
<u>Step-by-step explanation</u>:
Let the two consecutive odd integers be 1 and 3.
- The product of 1 and 3 is (1
3)=3 - The average of 1 and 3 is (1+3)/2 =4/2 = 2
- The square of their average is (2)² = 4
∴ The product 3 is one less than the square of their average 4.
Let the two consecutive even integers be 2 and 4.
- The product of 2 and 4 is (2
4)=8 - The average of 2 and 4 is (2+4)/2 =6/2 = 3
- The square of their average is (3)² = 9
∴ The product 8 is one less than the square of their average 9.
Thus, It is true that the product of two consecutive even integers are always one less than the square of their average.
Step-by-step explanation:
∑k=150(3k+2)
First find a1 and a50 :
a1=3(1)+2=5a20=3(50)+2=152
Then find the sum:
Sk=k(a1 + ak)2S50=50(5 + 152)2=3925
Since the basis is from year 1 to year 2, calculate first for the difference of their percentages. That would be:
Difference = year 2 - year 1
Difference = 2.32% - 1.1% = 1.22%
We apply this same value of percentage increase from year 2 to year. Thus, the percentage for year 3 is:
% Year 3 = % Year 2 + percentage increase
% Year 3 = 2.32% + 1.22%
% Year 3 = 3.54%