\(Q=\left(-1\right)+\left(-3\right)+\left(-5\right)+....+\left(-99\right)\)
\(Q=-\left(1+3+5+.....+99\right)\)
Áp dụng công thức tính dãy số ta có :
\(\frac{\left[\left(99-1\right):2+1\right].\left(99+1\right)}{2}=\frac{50.100}{2}=50.50=2500\)
=> Q = 2500
\(H=\frac{\left(1+97\right)\left(1+\frac{97}{2}\right)\left(1+\frac{97}{3}\right)\left(1+\frac{97}{4}\right)+...+\left(1+\frac{97}{99}\right)}{\left(1+99\right)\left(1+\frac{99}{2}\right)\left(1+\frac{99}{3}\right)\left(1+\frac{99}{4}\right)+...+\left(1+\frac{99}{97}\right)}\)
tinh nhanh:\(1+\left(-3\right)+5+\left(-7\right)+9+\left(11\right)+...+\left(-99\right)\)99)
a)\(\left(1\frac{1}{3}\right)\left(1\frac{1}{8}\right)\left(1\frac{1}{15}\right)...\left(1\frac{1}{99}\right)\)
b)\(\left(1+\frac{1}{3}\right)\left(1+\frac{1}{8}\right)\left(1+\frac{1}{15}\right)\left(1+\frac{1}{24}\right)\left(1+\frac{1}{99}\right)\)
Tính tổng sau:
\(1+\left(-2\right)+\left(3\right)+\left(-4\right)+5+.....+\left(-98\right)+99\)
bài 1 : tính
1) A = \(\left(\frac{1}{2}+1\right).\left(\frac{1}{3}+1\right).\left(\frac{1}{4}+1\right)........\left(\frac{1}{99}+1\right)\)
2) B = \(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right).\left(\frac{1}{4}-1\right).....\left(\frac{1}{99}-1\right)\)
1, Tính \(\frac{1}{2}-\left(\frac{1}{3}+\frac{2}{3}\right)+\left(\frac{1}{4}+\frac{2}{4}+\frac{3}{4}\right)-\left(\frac{1}{5}+\frac{2}{5}+\frac{3}{5}+\frac{4}{5}\right)+...+\left(\frac{1}{100}+\frac{2}{100}+\frac{3}{100}+...+\frac{99}{100}\right)\)2,Tính \(\left(1-\frac{1}{2^2}\right)x\left(1-\frac{1}{3^2}\right)x\left(1-\frac{1}{4^2}\right)x...x\left(1-\frac{1}{n^2}\right)\)
Tính:
\(B=1+\left(-3\right)+5+\left(-7\right)+.......+97+\left(-99\right)+101\)
A=\(\left(\frac{1}{2}+1\right)\left(\frac{1}{3}+1\right)\left(\frac{1}{4}+1\right)....\left(\frac{1}{99}+1\right)\)
b) \(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)\left(\frac{1}{4}-1\right)...\left(\frac{1}{100}-1\right)\)
\(F=1.100+2.\left(100-1\right)+3.\left(100-2\right)+...+\)\(99\left(100-98\right)+100\left(100-99\right)\)