Sum the series of n terms 1 + 3 + 7 + 15 + 31 .... to n terms?

Let -

Sn  =  1 + 3 + 7 + 15 + 31 ....  .....................+. Tn

Sn  =         1 + 3 +  7  +  15 +  31 +  .........    +  T(n-1)  + Tn
------------------------------------------------ ---------------------------------- Subtract
0  =  1 + 2 + 4 +  8  +  16 +  ............. .......................  -  Tn

=>  Tn   =   1 + [ 2 + 4 + 8 = 16 + .......... to (n-1) terms  ]

......................  2^n - 1
=>  Tn  =  1 * --------------   =   2^n  -  1
......... . ..............n - 1

Putting n = 1, 2, 3, 4, ........ n  Adding we get --

Sum =  ( 2 + 2^2 + 2^3 + ........ n Terms )

.....................  2^n - 1
=>  Sn  =  2 . ------------  -   n    =  2^n+1  - 2 - n  ...... Answer
.....................    2 - 1

1 + 3 + 7 + 15 + 31 + …… + the nth term
= (2 - 1) + (2² - 1) + (2³ - 1) + (2⁴ - 1) + (2⁵ - 1) + …… + (2ⁿ - 1)
= (2 + 2² + 2³ + 2⁴ + 2⁵ + …… + 2ⁿ) - n
= [n(n + 1)(2n + 1)/6] - (6n/6)
= (n/6) [(n + 1)(2n + 1) - 6]
= (n/6) (2n² + n + 2n + 1 - 6)
= n(2n² + 3n - 5)/6
= n(n - 1)(2n + 5)/6

Notice that it can all be written as

(2 - 1) + (2² - 1) + (2³ - 1) + (2⁴ - 1) + (2⁵ - 1) + …… + (2ⁿ - 1)

= (2 + 2² + 2³ + 2⁴ + 2⁵ + …… + 2ⁿ) - n


= 2(1 + 2 + 2² + 2³ + 2⁴ + 2⁵ + …… + 2^(n-1) ) - n

= 2* (2ⁿ - 1)/(2-1)  - n

= [last simplification for you].

Done!

Sₙ             =       1 + 3 + 7 + 15 + 31 ...  Xₙ          [Where the first term is X₁]
Sₙ + n       =       2 + 4 + 8 + 16 + 32 ... (Xₙ + 1)Sₙ + n + 1 = 1 + 2 + 4 + 8 + 16 + 32 ... (Xₙ + 1) By considering binary numbers, 1111 = 10000 - 1 and this pattern always holds. Then,Sₙ + n + 1 = 2ⁿ⁺¹ - 1Sₙ             = 2ⁿ⁺¹ - n - 2Sₙ = 2ⁿ⁺¹ - (n + 2)

1 , 3 , 7 , 15 , 31, .... 
an = 2^n - 1 
sum_(n=0)^n(2^n - 1) = -n + 2^(n + 1) - 2