The sum of 1 + 25 + 3(52) + 4(53) + .........upto n terms is
Given series, lets Sn = 1 + 25 + 352 + 453 + .........+n5n−1
15Sn = 15 + 252 + 353 + .........+n5n
Subtracting,
(1 - 15)Sn = 1 + 15 + 152 + 153 + .........+upto n terms - n5n
⇒ 45Sn = 1−15n45 - n5n ⇒ Sn = 2516 - 4n+516×5n−1