# Math Help - Arithmetic sequence problem

1. ## Arithmetic sequence problem

The sum of the first n terms of a certain progression is given by Sn= n^2 +3n
Find an expression for Sn-1 and deduce an expression for Un
I managed to figure out Un is 2n+2 but i cant quite get my head around the Sn-1 any help on this would be greatly appreciated

2. find S1 abd S2.

S1 = U1 i.e. first term.

S2 - S1 = U2.

U2 - U1 = d i.e. the common difference.

Now proceed.

3. thx but im still unsure sorry i feel realy thick right now i have the common diff d= 2 the first 4 terms are 4, 6, 8, 10 U1= 4 U2= 6 i still not sure what it wants when it asks for an expression for Sn-1

4. Originally Posted by pmh118
The sum of the first n terms of a certain progression is given by Sn= n^2 +3n
Find an expression for Sn-1 and deduce an expression for Un
I managed to figure out Un is 2n+2 but i cant quite get my head around the Sn-1 any help on this would be greatly appreciated
Hi pmh118,

$S_n=n^2+3n$

$S_{n-1}=(n-1)^2+3(n-1)=n^2-2n+1+3n-3=n^2+n-2$

$U_1+U_2+U_3+....+U_{n-1}=S_{n-1}$

$U_1+U_2+U_3+...+U_{n-1}+U_{n}=S_n$

$U_n=S_n-S_{n-1}=n^2+3n-n^2-n+2=2n+2$

5. Hello, pmh118!

$\text{The sum of the first }n\text{ terms of an arithmetic progression}$

. . $\text{is given by: }\;S_n \:=\: n^2 +3n$

$\text{(a) Find an expression for }S_{n-1}$

$\text{(b) Deduce an expression for }U_n$

$\text{I managed to figure out }U_n \:=\:2n+2}$ . . . . no

$\text{but I can't quite get my head around the }S_{n-1}$

(a) To find $S_{n-1}$, replace $\,n$ with $\,n-1$.

. . . $S_{n-1} \;=\;(n-1)^2 + 3(n-1) \;=\;n^2-2n+1+3n-3$

. . . $\boxed{S_{n-1} \;=\;n^2 + n -2}$

(b) The common difference $\,d$ is the difference of two consecutive terms.

. . . $d \;=\;S_n - S_{n-1} \;=\;(n^2 + 3n) - (n^2 + n - 2) \;=\;2n+2$

. . . $d \:=\:2(n+1)$ .[1]

. . . Sum of the first $\,n$ terms: . $S_n \;=\;\dfrac{n}{2}\bigg[2a + (n-1)d\bigg]$

. . . $\text{We have: }\;\dfrac{n}{2}\bigg[2a + (n-1)d\bigg] \;=\;n^2 +3n$

. . . Substitute [1]: . $\dfrac{n}{2}\bigg[2a + (n-1)2(n+1)\bigg] \;=\;n^2+3n$

. . . Solve for $\,a\!:\;\;a \;=\;4 + n - n^2$ .[2]

. . . The $\,n^{th}$ term is: . $U_n \;=\;a + (n-1)d$

. . . Substitute [1] and [2]: . $U_n \;=\;(4+n-n^2) + (n-1)2(n+1)$

. . . . . . . . . . . . . . . . . . . . $\boxed{U_n \;=\;n^2+n+2}$

6. Hi Soroban,

$S_n - S_{n-1}$ is not equal to d, but it is $U_n$

7. WoW Thanks all this has really helped me, I think it was the way i was multiplying out of the brackets i was missing the -2n from Sn-1 = (n – 1)2 +3(n – 1) = n2 –2n +1+3n – 3