100^2-99^2 = (100-99)x(100+99) =199
tương tự 98^2-97^2=195
=> cái bỉu thức trên thành
199+195+191+....+3
số số hạng: (199-3):4 + 1=50
tổng: [(199+3)*50]:2=5050
nếu thấy đúng thì k cho mình nha
= 100 x 100 - 99x99 - 98x98-...........-2x2-1x1
=(100-99-98-...-2-1) x (100-99-98-...-2-1)
=-4850x(-4850)
=23522500
Đặt :
\(A=100^2-99^2-98^2-........-2^2-1^2\)
\(\Leftrightarrow2A=2\left(2^{100}-2^{99}-........-2-1\right)\)
\(\Leftrightarrow2A=2^{101}-2^{100}-2^{99}-............-2^2-2\)
\(\Leftrightarrow2A-A=\left(2^{101}-2^{100}-.........-2^2-2\right)-\left(2^{100}-2^{99}-......-1\right)\)
\(\Leftrightarrow A=\left(2^{101}-2\right)\left(2^{100}-1\right)\)
\(\Leftrightarrow A=2^{101}-2^{101}+1=0+1=1\)
Gọi \(A=100^2-99^2-98^2-....-2^2-1^2\)
\(=100-\left(99^2+98^2+.....+2^2+1^2\right)\)
Đặt \(B=1^2+2^2+3^2+....+98^2+99^2\)
\(=1\left(2-1\right)+2 \left(3-1\right)+3\left(4-1\right)+....+99\left(100-1\right)\)
\(=1.2-1+2.3-2+3.4-3+....+99.100-99\)
\(=\left(1.2+2.3+3.4+....+99.100\right)-\left(1+2+3+...+99\right)\)
\(=\frac{99.100.101}{3}-\frac{99.100}{2}\)
\(=328350\)
\(\Rightarrow A=100^2-328350=-318350\)
ta có:
=100*100-99*99-..........-2*2-1*1
=[100-99]*[100-99]*.......*[2-1]*[2-1]
=1*1*......*1*1
=1.100 { vì ta có [100-1]:1+1=100[ số]}
=100
