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, [tex] a_{1} = 1 + 3(0) = 1+0 =1 [/tex]
similarly for next term plug n =1
For n = 1, [tex] a_{2} = 1 + 3(1) = 1+3 = 4 [/tex]
For n =2, [tex] a_{3} = 1 + 3(2) = 1+6 = 7 [/tex]
So on ,........and last term will be for n =4
For n =4, [tex] a_{5} = 1 + 3(4) = 1 + 12 = 13 [/tex]
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
[tex] S_{n} = \frac{n(a_{1} + a_{n})}{2} [/tex]
where n is total number of terms, [tex] a_{1} [/tex] and [tex] a_{n} [/tex] 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, [tex] a_{1} [/tex] as 1 and [tex] a_{5} [/tex] as 13 in sum formula
[tex] S = \frac{5(1 + 13)}{2} [/tex]
[tex] S = \frac{5(14)}{2} = \frac{70}{2} =35 [/tex]
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, [tex] a_{1} = 3(2) - 5 = 6 - 5 = 1 [/tex]
for i =3, [tex] a_{2} = 3(3) - 5 = 9 - 5 = 4 [/tex]
so on .... and for last term plug i as 6
for i =6, [tex] a_{5} = 3(6) - 5 = 18 - 5 = 13 [/tex]
So for sum we will plug n as 5 ( i =2,3,4,5,6 so total 5 terms), then plug [tex] a_{1} [/tex] as -2 and [tex] a_{5} [/tex] as 13 in sum formula
[tex] S = \frac{5(1 + 13)}{2} = \frac{5(14)}{2} = \frac{70}{2} =35 [/tex]
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
