Question

Sum 1 + 2a + 3a2 + 4a3 + .... to n terms.

Answer 1

Let, the sum of n terms of the given series be given by Sn   Sn=1+2a+3a2+4a3+⋯⋯⋯⋯⋯⋯⋯⋯⋯+nan−1          ⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯⋯(i)aSn=        a+2a2+3a3+⋯⋯⋯⋯⋯⋯⋯⋯⋯+(n−1)an−1+nan          ⋯⋯⋯⋯⋯⋯⋯⋯(ii)Now, subtracting equation (ii) from (i), we get:(1−a)Sn=1+a+a2+a3+ ⋯⋯⋯⋯⋯⋯⋯⋯+an−1−nan⇒(1−a)Sn=(1+a+a2+a3+ ⋯⋯⋯⋯⋯⋯⋯⋯+an−1)−nan⇒(1−a)Sn=an−1a−1−nan⇒(1−a)Sn=an−1−(a−1)nana−1=an−1−nan+1+nana−1=−nan+1+(n+1)an−1a−1⇒Sn=−nan+1+(n+1)an−1(a−1)(1−a)=nan+1−(n+1)an+1(a−1)2

Answer 2

Answer :

S=1+…+ka^(k-1)+…+na^(n-1)

s=1+..+a^k+…+a^(n-1)

+a[1+..a^(k-1)+…+a^(n-2)+…

+a^j[1+…+a^(k-1-j)+…+a^(n-2-j)

S=1-a^n/1-a+a*(1-a^n-1)/1-a +…a^k*(1-a^n-k)/1-a+…

S=1/1-a*[sum of (a^k -a^n)]k varies from 0 to n

S=1/1-a*sum a^k +1/1-a*sum a^n

s=1-a^n+1/(1-a)^2-n*a^n/1-a

$\rule{200}{2}$