Evaluate: ∑r=1∞(1+a+a2+…ar-1)r!
Simplifying the given expression:
∑r=1∞1+a+a2+…ar-1r!
Here we can see that numerator is a nth term G.P. with common ratio a and first term 1
Therefore,
⇒∑r=1∞1(ar-1)(a-1)r![UsingsumofnthtermofG.P.]⇒1a-1∑r=1∞arr!-∑r=1∞1r!⇒1a-1a1!+a22!+...-11!+12!+...[expanding]⇒1a-1ea-1-(e-1)[∑r=1∞arr!=ea-1and∑r=1∞1r!=e-1]⇒ea-ea-1
Therefore, ∑r=1∞(1+a+a2+…ar-1)r!=ea-ea-1
Sum of infinite terms A,AR,AR2,AR3,… is 15. If the sum of the squares of these terms is 150, then find the sum of AR2,AR4,AR6,….