Find the sum of the series `1+4+10+22+...`

Explanation

Let `S = 1 + 4 + 10 + 22 + .. + T_(n)`...(i)
`S = 1 + 4 + 10 + ... + T_(n -1) + T_(n)`....(ii)
`(i) - (ii) rArr T_(n) = 1 + (3 + 6 + 12 + .. + T_(n) - T_(n - 1))`
`T_(n) = 1 + 3 ((2^(n -1) - 1)/(2 - 1))`
`T_(n) = 3, 2^(n -1) - 2`
So `S_(n) = SigmaT_(n) = 3 Sigma 2^(n -1) - Sigma 2`
`= 3 ((2^(n) -1)/(2 -1)) - 2n = 3.2^(n) - 2n - 3`
Aliter Method :
Let `T_(n) = ar^(n) + b`, where r = 2
Now `T_(1) = 1 = ar + b`...(i)
`T_(2) = 4 = ar^(2) + b` ...(ii)
Solving (i) & (ii), we get
`a = (3)/(2) , b = -2`
`:. T_(n) = 3.2^(n - 1) - 2`
`rArr S_(n) = SigmaT_(n) = 3 Sigma 2^(n -1) - Sigma 2`
`= 3 ((2^(n) - 1)/(2 -1)) - 2n = 3.2^(n) - 2n -3`