Find the sum of the following series:
1 + 4/7 + 9/49 + 16/343 + .......... to infinity
The sum of the given series is S (say).
Then, we must have:
S = 1+ 4/7 + 9/(7^2) + 16/(7^3) +...... (i)
S/7 = 1/7 + 4/(7^2) + 9/(7^3) +..... (ii)
Subtracting (ii) from (i), we must have:
6*S 4-1 9-4 16-9
------ = 1 + ------- + -------- + --------- +.........
7 7 7^2 7^3
= 1 + 3/7 + 5/(7^2) + 7/(7^3)+.......
Then, we must have:
6*S 1 3-1 5 - 3 7 - 5
------ * (1 - -----) = 1 + -------- + ---------- + ----------
7 7 7 7^2 7^3
= 1 + 2{1/7 +1/(7^2) + 1/(7^3)+........}
36*S 2*(1/7) 1
=> ---------- = 1 + ---------------- = 1 + -----
49 1 - 1/7 3
=> S*(36/49) = 4/3
=> S = 49/27
Consequently, the required sum of the infinite series is 49/27
Edited on July 16, 2022, 11:17 pm
Puzzle Solution
