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\(50^2-49^2+48^2-47^2+...+2^2-1^2\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)\(=50+49+48+47+...+2+1\)
\(=\dfrac{50\left(50+1\right)}{2}\)
\(=\dfrac{50.51}{2}\)
\(=1275\)
\(D=\left(50^2+48^2+46^2+..+2^2\right)-\left(49^2+47^2+..+1^2\right)\)
\(=50^2+48^2+46^2+..+2^2-49^2-47^2-...-1^2\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=50+49+48+47+...+1\)
=\(\frac{\left(50+1\right)\cdot50}{2}=1275\)
Đặt A = 50^2 - 49^2 + 48^2 - 47^2 + ... + 2^2 - 1^2.
<=> A = (50 - 49)(50 + 49) + (48 - 47)(48 + 47) + ... + (2 - 1)(2 + 1)
= 99 + 95 + .. + 3
= (99 + 3)[(99 - 3) : 4 + 1] : 2 (cách tính tổng của dãy số cách đều)
= 1275.
(50^2+48^2+46^2+...+4^2+2^2)-(49^2+47^2+45^2+...+5^2+3^2+1^2)
=(50^2-49^2)+(48^2-47^2)+...+(2^2-1^2)
=(50+49)(50-49)+(48+47)(48-47)+...+(2+1)(2-1)=50+49+48+47+...+2+1
=\(\frac{50.51}{2}\)
=1275
=(502-492)+...+(22-12)
=(50-49)(50+49)+(48-47)(48+47)+...+(2-1)(2+1)
=1.99+1.95+1.91+...+1.3
=99+95+91+...+3
=(99+3)+(95+7)+...+
\(50^2-49^2+48^2-47^2+...+2^2-1^2\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(50+49\right)\left(50-49\right)+\left(48+47\right)\left(48-47\right)+...+\left(2+1\right)\left(2-1\right)\)
\(=99.1+95.1+...+3.1\)
\(=99+95+...+3\)
\(=3+...+95+99\)
Từ 3 đến 99 có: \(\left(99-3\right):4+1=25\left(\text{số hạng}\right)\)
Tổng là: \(\frac{\left(99+3\right)\times25}{2}=1275\)
\(C=50^2-49^2+48^2-47^2+...+2^2-1^2\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=99+95+...+3\)
\(=1275\)
Vậy C = 1275
\(C=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right).\)
\(C=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
\(C=50+49+48+....+2+1\)
\(C=\dfrac{\left(1+50\right).50}{2}=1275.\)
\(50^2-49^2+48^2-47^2+46^2-45^2+...+4^2-3^2+2^2\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+\left(46-45\right)\left(46+45\right)...+\left(4-3\right)\left(4+3\right)+4\)
\(=99+95+91+...+7+3+1\)
\(=\left(3+99\right).\left[\left(99-3\right):4+1\right]:2+1\)
\(=102x25:2+1=1276\)
Đặt \(THANG=50^2-49^2+48^2-47^2+....+2^2-1\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+....+\left(2^2-1\right)\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+....+\left(2-1\right)\left(2+1\right)\)
\(=50+49+48+47+....+2+1\)
\(=\dfrac{50\cdot\left(50+1\right)}{2}=\dfrac{50\cdot51}{2}=1275\)
(502+482+...+22) - (492+472+...+12)
= (502-492) + (482-472) + ... + (22-12)
= (50+49)(50-49) + (48+47)(48-47) + ... + (2+1)(2-1)
= 50+49+48+47+...+1
= \(\frac{\left(50+1\right).50}{2}=\frac{51.50}{2}=1275\)
\(\left(50^2+48^2+...+2^2\right)-\left(49^2+47^2+...+1^2\right)\)
\(=50^2+48^2+...+2^2-49^2-47^2-...-1^2\)
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
\(=\left(50-49\right)\left(50+49\right)+\left(48-47\right)\left(48+47\right)+...+\left(2-1\right)\left(2+1\right)\)
\(=50+49+48+47+...+2+1\)
\(=\dfrac{50\left(50+1\right)}{2}=\dfrac{50\cdot51}{2}=1275\)
Ta có : ( 502 + 482 + ... + 22 ) - ( 492 +472 + ... + 12 )
= 502 + 482 +...+ 22 - 492 -472 - 12
\(=\left(50^2-49^2\right)+\left(48^2-47^2\right)+...+\left(2^2-1^2\right)\)
= \(\left(50-49\right)\left(50+49\right)+\left(48-47\right)+..+\left(2-1\right).\left(2+1\right)\)
= \(50+49+48+47+...+2+1\)
= \(1257\)