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$$\sum_{k=1}^{n}\frac{k(k+1)}{2}p^{\frac{k(k-1)}{2}}\left(1-p^{k} \right)=?$$

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#1
dark templar

dark templar

    Kael-Invoker

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Bài toán: Cho trước số thực $p$ thuộc $[0;1]$.Hãy tính tổng :
$$S=\sum_{k=1}^{n}\frac{k(k+1)}{2}p^{\frac{k(k-1)}{2}}\left(1-p^{k} \right)$$
 
 


"Do you still... believe in me ?" Sarah Kerrigan asked Jim Raynor - Starcraft II:Heart Of The Swarm.

#2
dark templar

dark templar

    Kael-Invoker

  • Hiệp sỹ
  • 3788 Bài viết

Bài toán: Cho trước số thực $p$ thuộc $[0;1]$.Hãy tính tổng :
$$S=\sum_{k=1}^{n}\frac{k(k+1)}{2}p^{\frac{k(k-1)}{2}}\left(1-p^{k} \right)$$
 
 

 

Trường hợp $p=1$ và $p=0$ thì $S=0$,xét $p \neq 0$ và $p \neq 1$.

Để ý rằng :

$$p^{\frac{k(k-1)}{2}}(1-p^{k})=p^{\frac{k(k-1)}{2}}-p^{\frac{k(k+1)}{2}}=-\Delta p^{\frac{k(k-1)}{2}}$$

 

Nên theo SPTP thì:

$$-S=\left [ \frac{k(k+1)}{2}p^{\frac{k(k-1)}{2}} \right ]_{k=1}^{n+1}-\sum_{k=1}^{n}p^{\frac{k(k+1)}{2}}\Delta \frac{k(k+1)}{2}=\frac{(n+1)(n+2)}{2}p^{\frac{n(n+1)}{2}}-1-S_{1}$$

 

Biến đổi tổng $S_1$ như sau:

$$S_{1}=\sum_{k=1}^{n}p^{\frac{k(k+1)}{2}}\left ( \frac{(k+1)(k+2)}{2}-\frac{k(k+1)}{2} \right )=\sum_{k=1}^{n}(k+1)p^{\frac{k(k+1)}{2}}$$

 

Phân đoạn module $2$,ta sẽ có:

$$S_{1}=\sum_{i=1}^{\left \lfloor \frac{n}{2} \right \rfloor}p^{i}(2i+1)+2\sum_{j=1}^{\left \lfloor \frac{n+1}{2} \right \rfloor}p^{j}j=S_{2}+2S_{3}$$

 

Để ý rằng $\Delta p^{i}=(p-1)p^{i}$ nên:

$$S_{2}=\frac{1}{p-1}\sum_{i=1}^{\left \lfloor \frac{n}{2} \right \rfloor}\Delta p^{i}(2i+1)=\frac{1}{p-1}\left [ p^{i}(2i+1) \right ]_{i=1}^{\left \lfloor \frac{n}{2} \right \rfloor}-\frac{1}{p-1}\sum_{i=1}^{\left \lfloor \frac{n}{2} \right \rfloor}p^{i+1}\Delta (2i+1)$$

 

Do đó:

$$S_{2}=\frac{p^{\left \lfloor \frac{n}{2} \right \rfloor}(2\left \lfloor \frac{n}{2} \right \rfloor+1)-3p}{p-1}-\frac{2p}{(p-1)^{2}}\sum_{i=1}^{\left \lfloor \frac{n}{2} \right \rfloor}\Delta p^{i}$$

 

Hay:

$$S_{2}=\frac{p^{\left \lfloor \frac{n}{2} \right \rfloor}\left ( 2\left \lfloor \frac{n}{2} \right \rfloor +1 \right)-3p}{p-1}-\frac{2p^{2}(p^{\left \lfloor \frac{n}{2} \right \rfloor }-1)}{(p-1)^{2}}$$

 

Tương tự, ta có :

$$S_{3}=\frac{p^{\left \lfloor \frac{n+1}{2} \right \rfloor}\left \lfloor \frac{n+1}{2} \right \rfloor-p}{p-1}-\frac{p^{2}\left ( p^{\left \lfloor \frac{n+1}{2} \right \rfloor}-1 \right )}{(p-1)^2}$$

 

Do đó:

$$S_{1}=\frac{2\left \lfloor \frac{n}{2} \right \rfloor p^{\left \lfloor \frac{n}{2} \right \rfloor}+2\left \lfloor \frac{n+1}{2} \right \rfloor p^{\left \lfloor \frac{n+1}{2} \right \rfloor}+p^{\left \lfloor \frac{n}{2} \right \rfloor}-5p}{p-1}-\frac{2p^{2}}{(p-1)^{2}}\left ( p^{\left \lfloor \frac{n}{2} \right \rfloor}+p^{\left \lfloor \frac{n+1}{2} \right \rfloor} \right )$$

 

Vậy :

$$S=1+\frac{2\left \lfloor \frac{n}{2} \right \rfloor p^{\left \lfloor \frac{n}{2} \right \rfloor}+2\left \lfloor \frac{n+1}{2} \right \rfloor p^{\left \lfloor \frac{n+1}{2} \right \rfloor}+p^{\left \lfloor \frac{n}{2} \right \rfloor}-5p}{p-1}-\frac{2p^{2}}{(p-1)^{2}}\left ( p^{\left \lfloor \frac{n}{2} \right \rfloor}+p^{\left \lfloor \frac{n+1}{2} \right \rfloor} \right )-\frac{(n+1)(n+2)}{2}p^{\frac{n(n+1)}{2}}$$


"Do you still... believe in me ?" Sarah Kerrigan asked Jim Raynor - Starcraft II:Heart Of The Swarm.




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