Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có:
A=(100^2 -99^2)+(98^2 - 97^2)+(96^2 - 95^2)+.........+(2^2 - 1)
=(100-99)(100+99) + (98-97)(98+97) + (96-95)(96+95)+........+(2-1)(2+1)
=100+99+98+97+......+2+1=5050
Ở đây mình nhóm các hạng tử rồi AD hằng đẳng thức A^2 - B^2 = (A-B)(A+B)
\(A=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+...+2+1\)
\(=\left(100+1\right).\frac{100-1}{2}=\frac{101.99}{2}=\frac{9999}{2}\)
Đặt A = \(100^2-99^2+98^2-97^2+96^2-95^2+2^2-1^2\)
A\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\left(96-95\right)\left(96+95\right)+...+\left(2-1\right)\left(2+1\right)\)
A \(=199+195+191+...+3\)
A gồm \(\dfrac{\left(199-3\right)}{4}+1=50\) ( số hạng )
Vậy A = \(\dfrac{\left(199+3\right).50}{2}=5050\)
Đặt \(A=100^2-99^2+98^2-97^2+96^2-95^2+...+2^2-1^2\)
\(A=\left(100^2-99^2\right)+\left(98^2-97^2\right)+...+\left(2^2-1^2\right)\)
\(A=2.100-1+2.98-1+2.96-1+...+2.2-1\)
\(A=2.\left(100+98+...+2\right)-50\)
\(A=\dfrac{2.\left[\left(100-2\right):2+1\right].\left(100+2\right)}{2}-50\)
\(A=50.102-50\)
\(A=50.\left(201-1\right)\)
\(A=50.101\)
\(A=5050\)
Giải:
\(100^2-99^2+98^2-97^2+96^2-95^2+...+2^2-1^2\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+\left(96^2-95^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\left(96-95\right)\left(96+95\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=\left(100+99\right)+\left(98+97\right)+\left(96+95\right)+...+\left(2+1\right)\)
\(=100+99+98+97+96+95+...+2+1\)
\(=\dfrac{\left(100-1+1\right).\left(100+1\right)}{2}=5050\)
Vậy ...
Chúc bạn học tốt!
Ta có :
\(100^2-99^2+98^2-97^2+96^2-95^2+......+2^2-1^2\)
\(=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+\left(96-95\right)\left(96+95\right)+.....+\left(2-1\right)\left(2+1\right)\)
\(=100+99+98+97+96+95+......+2+1\)
\(=\dfrac{100.\left(100+1\right)}{2}=5050\)
\(=\left(100^2-99^2\right)+\left(98^2-97^2\right)+\left(96^2-95^5\right)+...+\left(2^2-1^2\right)\\ =\left(100-99\right).\left(100+99\right)+\left(98-97\right).\left(98+97\right)+\left(96-95\right).\left(96+95\right)+...+\left(2-1\right).\left(2+1\right)\\ =100+99+98+97+96+95+...+2+1\\ =50.101=5050\)
\(A=\left(100-99\right)\left(100+99\right)+\left(98-97\right)\left(98+97\right)+...+\left(2-1\right)\left(2+1\right)\)
\(A=100+99+98+97+...+2+1\)
\(A=\frac{100\cdot101}{2}=5050\)