Question

prove that Sn - Sn-1 = an

Answer 1

Answer:

$S_n = a_1 + a_2 +\cdots + a_n$

$S_{n-1} = a_1 + a_2 +\cdots +a_{n-1}$

So

$S_n - S_{n-1} = ( a_1 + a_2 +\cdots+ a_{n-1} + a_n ) - (a_1 + a_2 +\cdots + a_{n-1})\\= a_n, \quad\text{since all the other terms cancel}.$

Answer 2

Answer:

QED

Step-by-step explanation:

an = a + (n-1)d

Sn = (n/2)(a + a + (n-1)d)

Sn-1 = {(n-1)/2}(a + a + (n-2)d)

Sn - Sn-1 = (n/2)(a + a + (n-1)d)  - [  {(n-1)/2}(a + a + (n-2)d)]

Sn - Sn-1 = (n/2)(2a + (n-1)d)  - [  {(n-1)/2}(2a + (n-2)d)]

Sn - Sn-1 = na + n(n-1)d/2 - [  na - a  +(n-1)(n-2)d/2]

Cancelling na and splitting(n-1)(n-2)

Sn - Sn-1 = n(n-1)d/2 - [ - a  +n(n-1)d/2 -2(n-1)d/2]

Cancelling n(n-1)d/2

Sn - Sn-1 = - [ - a   -2(n-1)d/2]

Sn - Sn-1 = a + (n-1)d

Sn - Sn-1 = an

QED