Muhammad's recurrence

Using the recurrence relation given, all terms in the desired sum can be written in terms of $f_7$ and $f_8$ (or, really, any two consecutive terms). The easier terms to find are

$f_9=f_7+f_8$

and

$f_{10}=f_8+f_9=f_8+(f_7+f_8)=f_7+2f_8.$

Then, working "backwards," we can use the fact that $f_n=f_{n+2}-f_{n+1}$ for all integers $n\ge 1$ :

$\begin{aligned} f_6 &= f_8-f_7 \\ f_5 &= f_7-f_6 = f_7-(f_8-f_7) = 2f_7-f_8 \\ f_4 &= f_6-f_5 = (f_8-f_7)-(2f_7-f_8) = -3f_7+2f_8 \\ f_3 &= f_5-f_4 = (2f_7-f_8)-(-3f_7+2f_8) = 5f_7-3f_8 \\ f_2 &= f_4-f_3 = (-3f_7+2f_8)-(5f_7-3f_8) = -8f_7+5f_8 \\ f_1 &= f_3-f_2 = (5f_7-3f_8)-(-8f_7+5f_8) = 13f_7-8f_8 \end{aligned}$

So the sum of the ten terms now becomes

$\begin{aligned} &(13\!-\!8\!+\!5\!-\!3\!+\!2\!-\!1\!+\!1\!+\!0\!+\!1\!+\!1)f_7+(-8\!+\!5\!-\!3\!+\!2\!-\!1\!+\!1\!+\!0\!+\!1\!+\!1\!+\!2)f_8 \\ = &11f_7+0f_8 \\ = &11(56) \\ = &616. \end{aligned}$

Let $f_1=a$ and $f_2=b$ .

Then we have:

$f_3= a+b$

$f_4= a+2b$

$f_5= 2a+3b$

$f_6=3a+5b$

$f_7=5a+8b$

$f_8=8a+13b$

$f_9=13a+21b$

$f_{10}=21a+34b$

Then:

$f_1+f_2+f_3+f_4+f_5+f_6+f_7+f_8+f_9+f_{10}$

$=55a+88b=11*(5a+8b)=11*f_7=11*56=616$

Let $f_1$ be $a$ , and let $f_2$ be $b$ . Now, using the recurrence relation:

$f_3=a+b$ $f_4=a+2b$ $f_5=2a+3b$ $f_6=3a+5b$ $f_7=5a+8b$ $f_8=8a+13b$ $f_9=13a+21b$ $f_{10}=21a+34b$ Now, we are given $f_7=56=5a+8b$ , and we must find $f_1+f_2+f_3+f_4+f_5+f_6+f_7+f_8+f_9+f_{10}$ . Adding up our above terms, our sum is equal to $55a+88b$ . Finally, $55a+88b=11(5a+8b)=11f_7=11(56)=616$ .

Note that from the recurrence relation, that $f_3 = f_1+f_2$ . Similarly, $f_4 = f_3+f_2 = (f_1+f_2)+f_2 = f_1+2f_2$ . One can immediately see that a pattern emerges (easily proven by induction): that if $a_n$ is the nth Fibonacci number (and $n \geq 3$ ), then $f_n = a_{n-2}f_1+a_{n-1}f_2$ .

To prove this via induction, note that the base has already been previously shown. To prove the inductive step, assume that it holds for all integers up to some $k$ . Then we have that $f_{k+1} = f_{k-1} + f_k = (a_{k-3}f_1+a_{k-2}f_2) + (a_{k-2}f_1+a_{k-1}f_2)$ $= f_1(a_{k-3}+a_{k-2}) + f_2(a_{k-2}+a_{k-1}) = f_1a_{k-1} + f_2a_k$ , as desired.

Therefore, we can represent each $f_i$ in terms of $f_1$ and $f_2$ . Summing up the corresponding values for $i=1, 2, \ldots, 10$ yields that the sum is equivalent to $55f_1+88f_2$ . However, note that $f_7 = a_5f_1+a_6f_2 = 5f_1+8f_2=56$ . Therefore, we have that the sum is equal to $55f_1+88f_2 = 11(5f_1+8f_2) = 11f_7 = 11\cdot 56 = \boxed{616}$ .

suppose: f(1)=a and f(2)=b f(3) =a + b f(4) =a + 2b f(5) =2a + 3b f(6) =3a + 5b f(7) =5a + 8b f(8) =8a + 13b f(9) =13a + 21b f(10) =21a + 34b

Then sum f(1) until f(10) f(1)+f(2)+f(3)+f(4)+f(5)+f(6)+f(7)+f(8)+f(9)+f(10) = 55a+88b =11(5a + 8b) =11(56) =616

$f_1+f_2+f_3+f_4+f_5+f_6+f_7+f_8+f_9+f_{10}=$ $f_1+f_2+f_3+f_4+f_5+f_6+f_7+f_8+f_8+f_9+f_9=$ $f_1+f_2+f_3+f_4+f_5+f_6+f_7+f_7+f_7+f_8+f_8+f_8+f_8=$ $f_3+f_3+f_4+f_5+f_6+f_7+f_7+f_7+f_6+f_7+f_6+f_7+f_6+f_7+f_6+f_7=$ $f_3+f_5+f_5+f_6+f_6+f_6+f_6+f_6+f_7+f_7+f_7+f_7+f_7+f_7+f_7=$ $f_3+f_7+f_7+f_6+f_6+f_6+f_7+f_7+f_7+f_7+f_7+f_7+f_7=$ $f_3+f_6+f_6+f_6+f_7+f_7+f_7+f_7+f_7+f_7+f_7+f_7+f_7=$ $f_3+f_4+f_5+f_6+f_6+f_7+f_7+f_7+f_7+f_7+f_7+f_7+f_7+f_7=$ $f_5+f_5+f_6+f_6+f_7+f_7+f_7+f_7+f_7+f_7+f_7+f_7+f_7=$ $f_7+f_7+f_7+f_7+f_7+f_7+f_7+f_7+f_7+f_7+f_7=$ $56+56+56+56+56+56+56+56+56+56+56=$ 616

Denote $(x,y)$ to be $x*f_1 + y*f_2$ . Our strategy is to represent $f_7$ and the required sum.

By brute force, we calculate:

$f_1 = (1,0)$

$f_2 = (0,1)$

$f_3 = (1,1)$

$f_4 = (1,2)$

$f_5 = (2,3)$

$f_6 = (3,5)$

$f_7 = (5,8)$

$f_8 = (8,13)$

$f_9 = (13,21)$

$f_{10} = (21,34)$

We can find that $f_7 = (5,8)$ and the sum is $(55,88) = 11*(5,8) = 11*56 = 616$ as desired.

From the sequence, it is seen that the term is the sum of the previous and the previous previous term. This is a familiar sequence known as the fibonacci sequence where the original sequence goes 1,1,2,3,5,8...... Therefore, Let f1 = a Let f2 = b f3 = a +b f4 = a + 2b f5 = 2a + 3b f6 = 3a + 5b f7 = 5a + 8b f8 = 8a + 13b f9 = 13a + 21b

f10 = 21a + 34b

sum of the first 10 terms is 55a + 88b and f7 = 5a + 8b which is 11 times smaller than the sum. Therefore, if f7 = 56, then sum of the first 10 terms will be 56 x 11 = 616

Let $f_1=x$ and $f_2 =y$ . Then $f_7=5x+8y=56$ . The problem implies that $f_1+f_2+...+f_9+f_{10}$ is constant. So it doesn't matter what $x$ and $y$ are, as long as they satisfy $5x+8y=56$ . To make the calculations simpler, we let $x=8$ and $y=2$ . Calculating $f_1$ to $f_{10}$ and adding gives $616$ .

The sequence will be of the form : $\small {a, b, (a+b), (a+2b), (2a+3b), (3a+5b), \\ (5a+8b), (8a+12b), (13a+20b), (21a+32b)}$ The seventh term in this sequence is : $5a+8b$ Therefore $5a+8b=56$ The only solutions $\in \mathbb{Z^{+}}$ is $(0,7) , (8,2)$ So $a=0$ or $8$ and $b=7$ or $2$ $$ So in all there are two series possible. Substituting any solution set and summing the we get the sum as $\boxed{616}$

$$ Actually considering negative solutions as well, we get the same sum.

Observe that

$f_7 = f_6 + f_5 = 2f_5 + f_4 = 3f_4 + 2f_3 = 5f_3 + 3f_2 = 8f_2 + 5f_1.$

Then

$\begin{aligned} \sum_{i=1}^{10} f_i = & f_1 + f_2 + (f_1 + f_2) + (f_1 + 2f_2) + (2f_1 + 3f_2) +(3f_1 + 5f_2) \\ & \quad + (5f_1 + 8f_2) + (8f_1 + 13f_2) + (13f_1 + 21f_2) + (21f_1 + 34f_2) \\ = & 55f_1 + 88f_2 \\ = & 11f_7 \\ = & 616. \end{aligned}$

By simplifing we can say that f7=56=5f0+8f1

Which is just correct for f0=0 &f1=7

Or f0=10 &f1=2( neglected because fn+1>fn)

So 0,7,7,14,21,.....etc

I'm gonna right my answer in terms of rate of change. Look:

$f_{n} = f_{n-1} + f_{n-2}$ . I'm just gonna do this: $\frac{f_{n}}{f_{n-1}} = \frac{f_{n-1}}{f_{n-2}}$ .
$f_{n-2} = f_{n} - f_{n-1} \Longrightarrow \frac{f_{n}}{f_{n-1}} = \frac{f_{n-1}}{(f_{n} - f_{n-1})}$ .
This will give me: $\frac{f_{n}}{f_{n-1}} = 1 + \frac{f_{n-1}}{f_{n}}$ .

I will call $\frac{f_{n}}{n-1}$ as $\phi$ . This emply: $\phi = 1 + \frac{1}{\phi}$ . Multiply the terms by $\phi$ :
$\phi^{2} - \phi - 1 = 0 \Longrightarrow \phi = \frac{1 \pm \sqrt{5}}{2}$ . Since the sequence is made of consecutively sums, just consider $\phi = \frac{1 + \sqrt{5}}{2}$ , which is approximately $1.618$ .

Back to $f_{7} = 56$ . $\frac{56}{1.618} = 34.641$ . I'm gonna work with two integer answer for $f_{6}$ : $35$ or $34$ . Both will work. The rest of the job it's just elementary algebra with subtraction. Both answer will give this:

${0, 7, 7, 14, 21, 35, 56, 91, 147, 238}$ . Sum equal to $616$ ;
${8, 2, 10, 12, 22, 34, 56, 90, 146, 236}$ . Sum equal to $616$ .