Câu hỏi của RIBFUBUG

Question

Bài 1:

a, A=$2^2+3^2+4^2+5^2+....+100^2$

Lightning Farron · Accepted Answer

$A=2^2+3^2+4^2+5^2+...+100^2$

$A+1=1^2+2^2+3^2+4^2+5^2+...+100^2$

Ta có: $1^2+2^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}$

$A+1=\dfrac{100\cdot\left(100+1\right)\cdot\left(100\cdot2+1\right)}{6}$

$A+1=338350\Rightarrow A=338349$Câu trả lời: $A=2^2+3^2+4^2+5^2+...+100^2$

$A+1=1^2+2^2+3^2+4^2+5^2+...+100^2$

Ta có: $1^2+2^2+...+n^2=\dfrac{n\left(n+1\right)\left(2n+1\right)}{6}$

$A+1=\dfrac{100\cdot\left(100+1\right)\cdot\left(100\cdot2+1\right)}{6}$

$A+1=338350\Rightarrow A=338349$

Violympic toán 6