Question

Prove that a(n) = S(n)-S(n-1)​

Answer 1

Answer:

Step-by-step explanation:

S_n = a_1 + a_2 +.... + a_n

S_{n-1} = a_1 + a_2 +.....+a_{n-1}

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

Pl mark as brainliest

Answer 2

The image attached has an easy proof

the below one is a little bit detailed

choose the one which u feel comfortable my dear friend ✌✌

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