Matematické Fórum

stuart clark · 18. 02. 2020 15:21

If $S_{n}=\sum^{n}_{r=0}(-4)^r\cdot \binom{n+r}{2r}.$ Then $S_{2021}+2S_{2020}+S_{2019}=$

kerajs · 19. 02. 2020 16:25

$S_{n}=\sum^{n}_{r=0}(-4)^r \binom{n+r}{2r}=(-1)^n(2n+1)$

stuart clark · 21. 02. 2020 10:17

Thanks ↑ kerajs:

Can u please explain me

kerajs · 22. 02. 2020 16:15 — Editoval kerajs (22. 02. 2020 16:17)

$S_n=S_{n-1}-4\sum_{i=0}^{n-1}S_i \\ \Downarrow \\ S_{n}=-2S_{n-1}-S_{n-2} \ \ \wedge \ \ S_0=1 \ \ \wedge \ \ S_1=-3 \\ \Downarrow \\ S_n=(-1)^n(2n+1)$

stuart clark · 27. 02. 2020 15:24

Thanks ↑ kerajs:

but i did not understand how you calculate $\sum^{n-1}_{i=0}S_{i}$ and get the relation $S_{n}=-2S_{n-1}-S_{n-2}$

kerajs · 29. 02. 2020 07:17 — Editoval kerajs (29. 02. 2020 07:35)

$& {n \choose k} = {n-1 \choose k-1} + {n-1 \choose k} \\ & {n \choose n} = {n-1 \choose n-1} =...= {1 \choose 1}= {0 \choose 0}=1\\ & {n \choose 0} = {n-1 \choose 0} =...= {1 \choose 0}= {0 \choose 0}=1$

$& S_{n}=\sum^{n}_{i=0}(-4)^i \binom{n+i}{2i}=(-1)^n(2n+1)= \color{blue}{ {n \choose 0} }+\color{black}{\sum_{i=1}^{n-1}(-4)^i {n+i \choose 2i}} + \color{green}{ (-4)^n {2n \choose 2n} }=\\ & =\color{blue}{ {n-1 \choose 0} }+\color{black}{\sum_{i=1}^{n-1}(-4)^i \left[ {n+i -1\choose 2i-1}+{n+i -1\choose 2i}\right]} + \color{green}{ (-4)^n {2n-2 \choose 2n-2} }=\\ & = {n-1 \choose 0}+\sum_{i=1}^{n-1}(-4)^i {n+i -1\choose 2i}-4\left[ \sum_{i=1}^{n-1}(-4)^{i-1} {n+i -1\choose 2i-1}+(-4)^{n-1} {2n-2 \choose 2n-2} \right]=\\ & =S_{n-1}-4 \left[ \sum_{i=1}^{n-1}(-4)^{i-1} \left[ {n+i -2\choose 2i-2}+ {n+i -2\choose 2i-1}\right] +(-4)^{n-1} {2n-2 \choose 2n-2} \right]=\\ & =S_{n-1}-4 \left[S_{n-1}+ \sum_{i=1}^{n-1}(-4)^{i-1} {n+i -2\choose 2i-1}\right]=\\ & =S_{n-1}-4 \left[S_{n-1}+ \sum_{i=1}^{n-2}(-4)^{i-1} {n+i -2\choose 2i-1} + (-4)^{n-2}{2n -3\choose 2n-3}\right]=\\ & =S_{n-1}-4 \left[S_{n-1}+ \sum_{i=1}^{n-2}(-4)^{i-1} \left[{n+i -3\choose 2i-2}+{n+i -3\choose 2i-1} \right] + (-4)^{n-2}{2n -4\choose 2n-4}\right]=\\ & =S_{n-1}-4 \left[S_{n-1}+S_{n-2}+ \sum_{i=1}^{n-2}(-4)^{i-1} {n+i -3\choose 2i-1}\right]=\\ & =S_{n-1}-4 \left[S_{n-1}+S_{n-2}+ \sum_{i=1}^{n-3}(-4)^{i-1} {n+i -3\choose 2i-1}+(-4)^{n-3}{2n -5\choose 2n-5}\right]=\\ & =S_{n-1}-4 \left[S_{n-1}+S_{n-2}+ \sum_{i=1}^{n-3}(-4)^{i-1} \left[{n+i -4\choose 2i-2}+{n+i -4\choose 2i-1} \right] + (-4)^{n-3}{2n -6\choose 2n-6}\right]=\\ & =S_{n-1}-4 \left[S_{n-1}+S_{n-2}+S_{n-3}+ \sum_{i=1}^{n-3}(-4)^{i-1} {n+i -4\choose 2i-1}\right]=\\ & =....=\\ & =S_{n-1}-4 \left[S_{n-1}+S_{n-2}+S_{n-3}+...+S_2+ {2\choose 1} +(-4){3\choose 3} \right]=\\ & =S_{n-1}-4 \left[S_{n-1}+S_{n-2}+S_{n-3}+...+S_2+ {1\choose 0}+ {1\choose 1} +(-4){2\choose 2} \right]=\\ & =S_{n-1}-4 \left[S_{n-1}+S_{n-2}+S_{n-3}+...+S_2+S_1 + {1\choose 1} \right]=\\ & =S_{n-1}-4 \left[S_{n-1}+S_{n-2}+S_{n-3}+...+S_2+S_1 + {0\choose 0} \right]=\\ & =S_{n-1}-4 \left[S_{n-1}+S_{n-2}+S_{n-3}+...+S_2+S_1 + S_0 \right]=\\ &= S_{n-1}-4\sum_{i=0}^{n-1}S_i$

$& S_n=S_{n-1}-4\sum_{i=0}^{n-1}S_i=S_{n-1}-4S_{n-1}-4\sum_{i=0}^{n-2}S_i=-3S_{n-1}-4\sum_{i=0}^{n-2}S_i+(S_{n-2}-S_{n-2})=\\ & =-3S_{n-1}+S_{n-1}-S_{n-2}=-2S_{n-1}-S_{n-2}$

$& S_0=(-4)^0{0\choose 0}=1 \\ & S_1=(-4)^0{1\choose 0}+(-4)^1{2\choose 2}=-3 \\ & S_{n}=-2S_{n-1}-S_{n-2} \\ & r^2=-2r-1 \Rightarrow (r+1)^2=0 \\ & S_n=A(-1)^n+Bn(-1)^n \\ & \begin{cases} 1=A\cdot (-1)^0+B\cdot 0 \cdot (-1)^0 \ \text{ for } S_0 \\ -3=A\cdot (-1)^1+B\cdot 1 \cdot (-1)^1 \ \text{ for } S_1\end{cases} \\ & \begin{cases} A=1 \\ B=2 \end{cases} \\ & S_n=(-1)^n+2n(-1)^n =(-1)^n(2n+1)$