$S_k=\frac{a^k}{(a-b)(a-c)}+\frac{b^k}{(b-a)(b-c)}+\frac{c^k}{(c-a)(c-b)}$

V

vipboycodon

$S_k=\dfrac{a^k}{(a-b)(a-c)}+\dfrac{b^k}{(b-a)(b-c)}+\dfrac{c^k}{(c-a)(c-b)}$
Ta có: $S_0 = \dfrac{a^0}{(a-b)(a-c)}+\dfrac{b^0}{(b-a)(b-c)}+\dfrac{c^0}{(c-a)(c-b)}$
= $\dfrac{1}{(a-b)(a-c)}-\dfrac{1}{(a-b)(b-c)}+\dfrac{1}{(a-c)(b-c)}$
= $\dfrac{b-c}{(a-b)(a-c)(b-c)}-\dfrac{a-c}{(a-b)(b-c)(a-c)}+\dfrac{a-b}{(a-c)(b-c)(a-b)}$
= $0 $
$S_1$ và $S_2$ cũng làm như vậy và kq cũng $= 0$
 
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