Question

The sum Sn of the first n terms of a series is given by Sn= 1/4 (3n^2 + 13n). Find the nth term of the series and write down the first three terms of the series.
The sum Sn of the first n terms of a series is given by Sn= 1/4 (3n^2 + 13n). Find the nth term of the series and write down the first three terms of the series.
@Mathematics

Answer 1

n/2( a + a +(n-1)d) = n/2( 2a + nd - d) = (2an + n^2d -dn)/2
(4an + 2 n^2d - 2 dn)/4
(4a-2d)n + 2n^2d = 3n^2 + 13n

13 = 4a -2d
3 = 2d
d = 3/2

a = 4 = first term
second term is 11/2
third term 7

Answer 2

i still dont get it :S

Answer 3

Kind of hard to explain
It's like expanding the whole formula
You use the series formula and term formula

Answer 4

\[S_n=\frac{1}{4}(3n^2+13n)\]\[x_n=an+b\]\[S_n=\sum\limits_{k=1}^n x_k=\sum\limits_{k=1}^n (ak+b)=a\frac{n(n+1)}{2}+bn=\frac{a}{2}n^2+\left(\frac{a}{2}+b\right)n\]\[\frac{a}{2}=\frac{3}{4}\quad,\quad \frac{a}{2}+b=\frac{13}{4}\quad\Rightarrow\quad a=\frac{3}{2}\quad,\quad b=\frac{5}{2}\]\[x_n=\frac{1}{2}(3n+5)\quad;\quad x_1=4\quad,\quad x_2=11/2\quad,\quad x_3=7\]