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:
1.1,2,5,8
2.2,3,4,5
3.-7,10,15
Step-by-step explanation:
hope help
Answer:
p(t) = 0.19e0.10t
=>p'(t) = 0.19e0.10t (0.10*1)
=>p'(t) = 0.019e0.10t
t = 0 represents 1994
for 2002, t=2002-1994 =8
in 2002
average price =p(8)
=>average price = 0.19e0.10*8
=>average price =0.422853... million
rate of increase =p'(8)
=>rate of increase = 0.019e0.10*8
=>rate of increase =0.0422853... million per year
p(8)=$ 0.42 million
p'(8)=$ 0.042 million per year
Do you understand how to solve or do you want me to explain
