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Well known Fibonacci square series $(1)$

$$0^2+1^2+1^2+2^2+3^2+\cdots F_{n}^2=F_{n}F_{n+1}\tag1$$

$T_n=0,1,3,6,10,..$ and $F_n=0,1,1,2,3,...$ For $n=0,1,2,3,...$

Now we included Triangular numbers into $(1)$ as shown below

$$T_0F_0^2=F_1F_2-1$$

$$T_1F_0+T_0F_1=F_2F_3-1$$

$$T_2F_0^2+T_1F_1^2+T_0F_2^2=F_3F_4-2$$

$$T_3F_0^2+T_2F_1^2+T_1F_2^2+T_0F_3^2=F_4F_5-2$$

$$T_4F_0^2+T_3F_1^2+T_2F_2^2+T_1F_3^2+T_0F_4^2=F_5F_6-3$$

$$T_5F_0^2+T_4F_1^2+T_3F_2^2+T_2F_3^2+T_1F_4^2+T_0F_5^2=F_6F_7-3$$

Observing the series involving Triangular and Fibonacci numbers together we found the following closed form.

For even terms

$$\sum_{k=0}^{2n+1}T_{2n+1-k}\color{red}{F_k^2}=F_{2n+1}F_{2n+2}-n-1\tag2$$

For odd terms

$$\sum_{k=0}^{2n}T_{2n-k}\color{red}{F_k^2}=F_{2n}F_{2n+1}-n\tag3$$

How can we prove $(2)$ and $(3)$?

An attempt:

Knowing that $T_n={n(n+1)\over 2}$ then $(3)$ becomes

$${1\over 2}\sum_{k=0}^{2n}(2n-k)(2n-k+1)F_k^2=F_{2n}F_{2n+1}-n$$

Simplified down to

$$(4n^2+2n)\sum_{k=0}^{2n}F_k^2+\sum_{k=0}^{2n}(k^2-k-4nk)F_k^2=2F_{2n}F_{2n+1}-2n$$

finally down to

$$\sum_{k=0}^{2n}(k^2-k-4nk)F_k^2=(2-2n-4n^2)F_{2n}F_{2n+1}-2n$$

we are not sure what to do next...

  • 0
    Have you tried $(F_n)^2=F_{2n-1}-(F_{n-1})^2$? the sum over linear Fibonaccis times triangular number should be doable by differentiating the generating function of the Fibonaccis – 2017-02-06
  • 0
    Let us consider the generating function of the squared Fibonacci numbers $F_2(x)$: $$ F_2(x)=\sum_{l=0}^n F_l^2x^l=\sum_{l=0}^n (F_{2l-1}-F_{l-1}^2)x^l=...=\sum_{l=0}^n \sum_{r=1}^lF_{2r-1}x^l=\sum_{l=0}^nF_{2l}x^l=\frac{1}{2}(F_1(x)+F_1(-x)) $$ where $F_1(x)=\frac{x}{1-x-x^2}$ is generating function of the usual Fibonacci numbers and i used several standard identities found at https://en.wikipedia.org/wiki/Fibonacci_number. All what is left now is to combine proper derivatives of $F_2(x)$ since $T_{n}=n(n+1)/2$ and setting $x=1$ in the end. – 2017-02-06

1 Answers 1

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This answer uses $$\sum_{k=0}^{n}F_k^2=F_nF_{n+1}\tag4$$ $$\sum_{k=0}^{n}kF_k^2=nF_nF_{n+1}-F_n^2+\frac{1+(-1)^{n-1}}{2}\tag5$$ $$(-1)^n=F_{n-1}F_{n+1}-F_n^2\tag6$$ The proofs are written at the end of the answer.

We want to prove that $$\sum_{k=0}^{m}T_{m-k}F_k^2=F_{m}F_{m+1}-\left\lceil\frac{m}{2}\right\rceil\tag7$$

Let us prove $(7)$ by induction on $m$ using $(4)(5)(6)$.

$(7)$ holds for $m=1$.

Supposing that $(7)$ holds for some $m\ (\ge 1)$ gives $$\begin{align}\sum_{k=0}^{m+1}T_{m+1-k}F_k^2&=\sum_{k=0}^{m}(T_{m-k}+m+1-k)F_k^2\\\\&=\left(\sum_{k=0}^{m}T_{m-k}F_k^2\right)+\left(\sum_{k=0}^{m}(m+1-k)F_k^2\right)\\\\&=F_{m}F_{m+1}-\left\lceil\frac m2\right\rceil+(m+1)\left(\sum_{k=0}^{m}F_k^2\right)-\left(\sum_{k=0}^{m}kF_k^2\right)\\\\&=F_{m}F_{m+1}-\left\lceil\frac m2\right\rceil+(m+1)F_{m}F_{m+1}-\left(mF_{m}F_{m+1}-F_{m}^2+\frac{1+(-1)^{m-1}}{2}\right)\\\\&=2F_{m}F_{m+1}+F_{m}^2-\left\lceil\frac m2\right\rceil-\frac{1+(-1)^{m-1}}{2}\\\\&=2F_{m}F_{m+1}+F_{m-1}F_{m+1}-(-1)^m-\left\lceil\frac m2\right\rceil-\frac{1+(-1)^{m-1}}{2}\\\\&=F_{m+1}(F_m+F_m+F_{m-1})-(-1)^m-\left\lceil\frac m2\right\rceil-\frac{1+(-1)^{m-1}}{2}\\\\&=F_{m+1}F_{m+2}-(-1)^m-\left\lceil\frac m2\right\rceil-\frac{1+(-1)^{m-1}}{2}\\\\&=F_{m+1}F_{m+2}-\left\lceil\frac{m+1}{2}\right\rceil\qquad\blacksquare\end{align}$$

Let us prove $(4)$ by induction on $n$. $$\sum_{k=0}^{n}F_k^2=F_nF_{n+1}\tag4$$

$(4)$ holds for $n=1$.

Supposing that $(4)$ holds for some $n\ (\ge 1)$ gives $$\sum_{k=0}^{n+1}F_k^2=F_nF_{n+1}+F_{n+1}^2=F_{n+1}(F_n+F_{n+1})=F_{n+1}F_{n+2}\qquad \blacksquare$$

Next, let us prove $(5)$ by induction on $n$ using $(6)$.

$$\sum_{k=0}^{n}kF_k^2=nF_nF_{n+1}-F_n^2+\frac{1+(-1)^{n-1}}{2}\tag5$$ $$(-1)^n=F_{n-1}F_{n+1}-F_n^2\tag6$$

$(5)$ holds for $n=1$.

Supposing that $(5)$ holds for some $n\ (\ge 1)$ gives $$\begin{align}\sum_{k=0}^{n+1}kF_k^2&=nF_nF_{n+1}-F_n^2+\frac{1+(-1)^{n-1}}{2}+(n+1)F_{n+1}^2\\\\&=nF_nF_{n+1}+nF_{n+1}^2+F_{n+1}^2-F_n^2+\frac{1+(-1)^{n-1}}{2}\\\\&=nF_{n+1}(F_n+F_{n+1})+(F_{n+1}+F_n)(F_{n+1}-F_n)+\frac{1+(-1)^{n-1}}{2}\\\\&=nF_{n+1}F_{n+2}+F_{n+2}F_{n+1}-F_{n+2}F_n+\frac{1+(-1)^{-1}}{2}\\\\&=(n+1)F_{n+1}F_{n+2}-(F_{n+1}^2+(-1)^{n+1})+\frac{1+(-1)^{n-1}}{2}\\\\&=(n+1)F_{n+1}F_{n+2}-F_{n+1}^2+\frac{1+(-1)^n}{2}\end{align}$$

Finally, let us prove $(6)$ by induction on $n$.

$$(-1)^n=F_{n-1}F_{n+1}-F_n^2\tag6$$

$(6)$ holds for $n=1$.

Supposing that $(6)$ holds for some $n\ (\ge 1)$ gives $$\begin{align}(-1)^{n+1}&=-(-1)^n\\\\&=-F_{n-1}F_{n+1}+F_n^2\\\\&=-(F_{n+1}-F_n)F_{n+1}+F_n^2\\\\&=-F_{n+1}^2+F_nF_{n+1}+F_n^2\\\\&=-F_{n+1}^2+F_n(F_{n+1}+F_n)\\\\&=F_{n}F_{n+2}-F_{n+1}^2\qquad\blacksquare\end{align}$$