Finding the sum upto n terms of the sequence ...:?

The exact formula for the nth term of the Fibonacci sequence

1, 1, 2, 3, 5, 8, ...

is

[(1+√5)^n - (1-√5)^n] / [(2^n)√5]

= (1/√5)[r^n - s^n]

where r = (1+√5)/2 and s = (1-√5)/2

However in this case the first 1 is omitted, so we need to replace n by n+1, making the term

(1/√5)[r^(n+1) - s^(n+1)]

Summing this series amounts to summing two geometric series.

The sum to n terms of the series ar^(n-1) is

a(r^n - 1)/(r - 1)

hence the sum of ar^(n+1) is

a(r^2)(r^n - 1)/(r - 1)

and so the sum of the given series is

(1/√5)[ (r^2)(r^n - 1)/(r - 1) - (s^2)(s^n - 1)/(s - 1)] ............... (1)

Now (r^2)(r^n - 1)/(r - 1)

= [([(1+√5)/2]^(n+2) - [(1+√5)/2]^2)/((1+√5)/2 - 1)

= [([(1+√5)/2]^(n+2) - [(1+√5)/2]^2)/((-1+√5)/2)

= - [([(1+√5)/2]^(n+2) - [(1+√5)/2]^2)/((1-√5)/2)

To get the corresponding expression for

a(s^n - 1)/(s - 1)

replace √5 by -√5:

- [([(1-√5)/2]^(n+2) - [(1-√5)/2]^2)/((1+√5)/2)

The common denominator of these two expressions is

((1-√5)/2)*((1+√5)/2)

= -1

Hence

(r^(n+2) - 1)/(r - 1) - (s^(n+2) - 1)/(s - 1)

= [([1+√5)/2]^(n+3) - ((1+√5)/2)^3 - [([1-√5)/2]^(n+3) - ((1-√5)/2)^3]

= [([1+√5)/2]^(n+3) - [([1-√5)/2]^(n+3) - 2√5)

Now multiply by 1/√5 to complete the evaluation of expression (1) above:

S(n) = [([(1+√5)/2]^(n+3) - [([(1-√5)/2]^(n+3)) / √5] - 2

....... = [(1+√5)^(n+3) - (1-√5)^(n+3))] /[(2^(n+3))√5] - 2

Looking slightly less awkward (perhaps!):

S(n) = [(1+√5)^(n+3) - (1-√5)^(n+3))]

.......... _______________________ .... - 2

...................... (2^(n+3))√5

To check, note that

S(1) = [(1+√5)^4 - (1-√5)^4)] /[(2^4)√5] - 2

....... = 48√5 / (16√5) - 2

....... = 1

S(2) = [(1+√5)^5 - (1-√5)^5)] /[(2^5)√5] - 2

....... = 160√5 / (32√5) - 2

....... = 3

S(3) = [(1+√5)^6 - (1-√5)^6)] /[(2^6)√5] - 2

....... = 512√5 / (64√5) - 2

....... = 6

confirming that we have the correct formula.

EDIT: Notice that Dakota's answer applies to the standard series I stated at the beginning. The series you presented omits the first 1, hence each sum is 1 less than the corresponding sum of the standard series, and that is why my formula subtracts 2.

FURTHER EDIT: Dakota's quoting of the result from Wikipedia led me to have a look at the most general Fibonacci sequence, whose n'th term is

u(n) = ar^n + bs^n where

r = (1+√5)/2 and s = (1-√5)/2

[Applying the information u(1) = 1 = u(2) determines the values for the standard Fibonacci sequence

a = 1/√5; b = -1/√5]

Then I calculated the sum up to the n'th term as

S(n) = ar^(n+2) + bs^(n+2) - ar^2 - bs^2

....... = u(n+2) - u(2)

So to find the sum up to the n'th term, calculate the (n+2)'th term and subtract the 2nd term.

This agrees with Dakota's (Wikipedia's) formula for the standard sequence, where u(2) = 1;

but in the sequence you asked about u(2) = 2, so we subtract 2, as I found already.

"it may be shown that the sum of the first Fibonacci numbers up to the nth is equal to the n+2nd Fibonacci number minus 1"

So you'd add 2 to n, calculate 1.6180)^n - (-0.618)^n]/sqrt(5) and subtract 1.

Hy's answer gives the right sums, so it must be right, but why is the exponent n+3 when the general formula has n+2 in it?