Question

The sum of the first n terms of a series is S sub n=2n^2

a. Find term 1, term 2, and term 3.
b. Find S sub n - S sub (n-1)

Answer 1

For the sequence T1, T2, T3, ....

The partial sums S1, S2, S3, ... are defined as follows

S1 = T1

S2 = T1 + T2

S3 = T1 + T2 + T3

etc etc


So let's find S1, S2, S3


S1 = 2*(1)^2

S1 = 2*1

S1 = 2


S2 = 2*(2)^2

S2 = 2*4

S2 = 8


S3 = 2*(3)^2

S3 = 2*9

S3 = 18


So S1, S2, and S3 are


S1 = 2,
S2 = 8,
S3 = 18


Since S1 = T1, this means that T1 = 2


Because S2 = T1 + T2, we know that

S2 = T1 + T2

8 = 2 + T2

8 - 2 = T2

6 = T2

T2 = 6


and finally, since S3 = T1 + T2 + T3, we can say

S3 = T1 + T2 + T3

18 = 2 + 6 + T3

18 = 8 + T3

18 - 8 = T3

10 = T3

T3 = 10


So the first three terms are 2, 6, and 10


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S[n] = 2n^2


S[n-1] = 2*(n-1)^2



So


S[n] - S[n-1] = 2n^2 - 2(n-1)^2


S[n] - S[n-1] = 2n^2 - 2(n^2 - 2n + 1)


S[n] - S[n-1] = 2n^2 - 2n^2 + 4n - 2


S[n] - S[n-1] = 4n - 2