Question

[Vui lòng điền đầy đủ]

tính \(s=1^2+2^2+3^2+...+99^2+100^2\)

Answer 1

[⭐Hannie⭐]

Tham khảo

S =1² + 2² + 3² +......+ 99² + 100²

   = 1 + 2.(1 + 1) + 3.(1 + 2) + ... + 99.(1 + 98) + 100.(99 + 1)

   = 1 + 2.1 + 2 + 3.1 + 3.2 +... + 99.1 + 99.98 + 100.99 + 100.1

    = (2.1 + 2.3 + ... + 99.99 ) + (1 + 2 + 3 + ... + 99  + 100)

   = 333300 + 5050

   = 338350

Answer 2

[Nguyen My Van]

\(S=1^2+2^2+3^2+...+99^2+100^2\)

\(=1.1+2.2+3.3+...+100.100\)

\(=1\left(2-1\right)+2\left(3-1\right)+3\left(4-1\right)+...+100\left(101-1\right)\)

\(=\left[1.2-1+2.3-1.1+3.4-3+1+...+100.101-100.1\right]\)

\(=\left[1.2+2.3+3.4+...+100.101\right]-\left(1+2+3+...+100\right)\)

\(=\dfrac{100.101.102}{3}-\dfrac{100.101}{2}\)

\(\dfrac{100.101.\left(2.100+1\right)}{6}=338350\)