Taking first summation ∑(1+3n) first
First see few terms of this series. To get first term plug n =0 in 1+3n
So for n =0,
similarly for next term plug n =1
For n = 1,
For n =2,
So on ,........and last term will be for n =4
For n =4,
So we get terms as 1,4,7,.. with constant difference of 3 between succesive terms. So its arithemetic series. Sum formula for arithmetic series is
where n is total number of terms, and are first and last term in series
So here n = 5 ( as n is from 0 till 4). so plug value of n as 5, as 1 and as 13 in sum formula
So sum of first series is 35
Similarly find sum of second series. Plug i values in 3i-5 as shown and find few terms in series
for i =2,
for i =3,
so on .... and for last term plug i as 6
for i =6,
So for sum we will plug n as 5 ( i =2,3,4,5,6 so total 5 terms), then plug as -2 and as 13 in sum formula
So sum of second series is also 35
So difference between sums of both series will be 35-35 =0
Final answer here is 0
2. x- -3, -1, 2, 5, 7 y- 9, 5, 4, -5, -7
Answer:
1. Domain: (3, 5, 7, 8, 11)
Range: ( 6, 7, 9, 14)
2. Domain: (-3, -1, 2, 5, 7)
Range: (-7, -5, -4, 5, 9)
Step-by-step explanation:
The domain is the set of all x values in the function. The range is the set of all y values in the function. It is written in parenthesis from least to greatest. Do not repeat values even if the function has them more than once.
1. Domain: (3, 5, 7, 8, 11)
Range: ( 6, 7, 9, 14)
2. Domain: (-3, -1, 2, 5, 7)
Range: (-7, -5, -4, 5, 9)