Answer to Question #122820 in Math for Ojugbele Daniel

Question

The sum of the first n terms of a series is 2n^2 -2. Find the nth term and show that the series is an arithmetic progression.

Answer 1

$S_n=2n^2-n$

$S_1=2(1)^2-1=1=>a_1=1$ $S_2=S_1+a_2=1+a_2=2(2)^2-2=6=>a_2=5$ $S_3=S_2+a_3=6+a_3=2(3)^2-3=15=>a_3=9$ $S_4=S_3+a_4=15+a_4=2(4)^2-4=28=>a_4=13$ $S_5=S_4+a_5=28+a_5=2(5)^2-5=45=>a_5=17$ $...$ $S_{n+1}=S_n+a_{n+1}=2(n)^2-n+a_{n+1}=$ $=2(n+1)^2-(n+1)=>a_{n+1}=4n+1$

$a_{n+1}=4n+1$

$a_n=1+4(n-1)$

$d=a_{n+1}-a_n=4$

The series is an arithmetic progression: $a_1=1, d=4.$

$a_1=a_1+d(n-1)=1+4(n-1)$

$S_n={2a_1+d(n-1)\over 2}\cdot n=n+2n(n-1)=2n^2-n$