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[Dark Magician 2k2]

Tình các tổng sau

a) $S_{1}=\sum_{k=0}^{1008}C^{2k}_{2017}$

b) $S_{2}=\sum_{k=0}^{504}C^{4k}_{2017}$

c) $S_{3}=\sum_{k=0}^{2017}(C^{k}_{2017})^2$

d) $S_{4}=\sum_{k=0}^{2016}k^2.C^{k}_{2016}$

[Quoc Tuan Qbdh]

Tình các tổng sau

a) $S_{1}=\sum_{k=0}^{1008}C^{2k}_{2017}$

b) $S_{2}=\sum_{k=0}^{504}C^{4k}_{2017}$

c) $S_{3}=\sum_{k=0}^{2017}(C^{k}_{2017})^2$

d) $S_{4}=\sum_{k=0}^{2016}k^2.C^{k}_{2016}$

a) Ta có:

$$(1+1)^{2017}=C^{0}_{2017}+C^{1}_{2017}+C^{2}_{2017}+C^{3}_{2017}+C^{4}_{2017}+....+C^{2017}_{2017}$$

$$(1-1)^{2017}=C^{0}_{2017}-C^{1}_{2017}+C^{2}_{2017}-C^{3}_{2017}+C^{4}_{2017}+....-C^{2017}_{2017}$$

Suy ra, $S_1=\dfrac{2^{2017}}{2}=2^{2016}$

 

b) Ta có:

$$(1+1)^{2017}=C^{0}_{2017}+C^{1}_{2017}+C^{2}_{2017}+C^{3}_{2017}+C^{4}_{2017}+....+C^{2017}_{2017}$$

$$(1-1)^{2017}=C^{0}_{2017}-C^{1}_{2017}+C^{2}_{2017}-C^{3}_{2017}+C^{4}_{2017}+....-C^{2017}_{2017}$$

$$(1+i)^{2017}=C^{0}_{2017}+C^{1}_{2017}.i-C^{2}_{2017}-C^{3}_{2017}.i+C^{4}_{2017}+....+C^{2017}_{2017}.i$$

$$(1-i)^{2017}=C^{0}_{2017}-C^{1}_{2017}.i-C^{2}_{2017}+C^{3}_{2017}.i+C^{4}_{2017}+....-C^{2017}_{2017}.i$$

Suy ra, $S_2=\dfrac{2^{2017}+(1+i)^{2017}+(1-i)^{2017}}{4}$

Mặt khác, theo hệ thức Moivre:
$$(1+i)^{2017}=(\sqrt{2})^{2017}\left(cos\dfrac{\pi}{4}+i.sin\dfrac{\pi}{4}\right)^{2017}$$

$$=(\sqrt{2})^{2017}.\left(cos\dfrac{2017.\pi}{4}+i.sin\dfrac{2017.\pi}{4}\right)=2^{1008}(1+i)$$

$$(1-i)^{2017}=(\sqrt{2})^{2017}\left(cos\dfrac{-\pi}{4}+i.sin\dfrac{-\pi}{4}\right)^{2017}$$

$$=(\sqrt{2})^{2017}.\left(cos\dfrac{-2017.\pi}{4}+i.sin\dfrac{-2017.\pi}{4}\right)=2^{1008}(1-i)$$

Do đó, $S_2=2^{2015}+2^{1007}$.

 

c) Ta có:
$$(1+x)^{2017}.(1+x)^{2017}=(C^{0}_{2017}+C^{1}_{2017}.x+....+C^{2016}_{2017}.x^{2016}+C^{2017}_{2017})(C^{2017}_{2017}+C^{2016}_{2017}.x^{2016}+....+C^{1}_{2017}.x+C^{0}_{2017})$$

$$(1+x)^{4034}=C^{0}_{4034}+....+C^{2017}_{4034}.x^{2017}+....+C^{4034}_{4034}.x^{4034}$$

Đồng nhất hệ số của $x^{2017}$ ta được: $S_3=C^{2017}_{4034}$

 

d) Xét số hạng tổng quát:

$$k^{2}.C^{k}_{2016}=k^{2}.\dfrac{2016!}{(2016-k)!.k!}=\dfrac{2016!.(k-1+1)}{(2016-k)!.(k-1)!}=2016.2015.\dfrac{2014!}{(2016-k)!.(k-2)}+2016.\dfrac{2015!}{(2016-k)!(k-1)!}=2016.2015.C^{k-2}_{2014}+2016.C^{k-1}_{2015}$$

Đến đây thay $k=2,...,2016$ để ý $\sum_{k=0}^{n} C^{k}_{n}=(1+1)^{n}=2^{n}$.