If un=a+bn+c2n, what's the sum of the first n terms?
Question
The nth term of a certain series is of the form a+bn+c2n, where a, b, and c are numbers. If the first three terms are 2, −1, and 3, find the values of a, b, and c, and the sum of the first n terms.
Solution
We have
un=a+bn+c2n,
so
a+b+2ca+2b+4ca+3b+8c=2,=−1,=3.
Taking (2) − (1) gives b+2c=−3, while (3) − (2) gives b+4c=4.
Subtracting these two equations gives 2c=7, and so c=27.
Substituting back into the earlier equations gives b=−10, and a=5.
So the nth term un=5−10n+272n=5−10n+7×2n−1.
Check for the third term: does 3=5−30+28? Yes.
… and the sum of the first n terms.
This will be
(a+b+2c)+(a+2b+4c)+(a+3b+8c)+...+(a+nb+2nc)=na+b(1+2+..+n)+c(2+22+23+...+2n)=na+b2n(n+1)+2c(2n−1)=5n−5n2−5n+7×2n−7=−5n2+7×2n−7.
Check: when n=3, the sum of the first n terms is 4. Our formula gives −45+56−7=4, so we have agreement.