Answer:
The given expanded sum of the series is 
Step-by-step explanation:
Given problem can be written as
To find their sums:
Now expanding the series
That is put n=5,6,7,8,9 in the given summation
![\sum\limits_{n=5}^{9}3n+2=[3(5)+2]+[3(6)+2]+[3(7)+2]+[3(8)+2]+[3(9)+2]](https://tex.z-dn.net/?f=%5Csum%5Climits_%7Bn%3D5%7D%5E%7B9%7D3n%2B2%3D%5B3%285%29%2B2%5D%2B%5B3%286%29%2B2%5D%2B%5B3%287%29%2B2%5D%2B%5B3%288%29%2B2%5D%2B%5B3%289%29%2B2%5D)
![=[15+2]+[18+2]+[21+2]+[24+2]+[27+2]](https://tex.z-dn.net/?f=%3D%5B15%2B2%5D%2B%5B18%2B2%5D%2B%5B21%2B2%5D%2B%5B24%2B2%5D%2B%5B27%2B2%5D)
(adding the terms)
Therefore
Therefore the given sum of the series is
The given expanded sum of the series is
Answer:
24
Step-by-step explanation:
at (4, 8): 8 = 4k ⇒ k = 2
So 
when 
, 
The answer is 11abc^2 + 12a^2b + 18b^2c because:
14abc^2-3abc^2 is 11abc^2
12a^2b + 0 = 12a^2b
16b^2c +2b^2c = 18b^c
Answer:
D.
Step-by-step explanation:
It just is hard to explain but I had the same question.
B. Abigail used less than Carly and the total was 109.
