HAKED BY PAKISTAN 2011
Tính tổng
A=\(1^3+2^3+3^3+...+100^3\)
B=\(2^3+4^3+...+98^3\)
C=\(1^3+3^3+5^3+...+99^3\)
D=\(1^3-2^3+3^3-4^3+...+99^3-100^3\)
Tính A= 2^3+ 4^3+ 6^3+ ...+98^3+ 100^3
B= 1^3+ 3^3+ 5^3+ ...+97^3+ 99^3
C= 1^3+ 2^3- 3^3+ 4^3- 5^3+ 6^3- ...+98^3- 99^3+ 100^3
Tính A= 2^3+ 4^3+ 6^3+ ...+98^3+ 100^3
B= 1^3+ 3^3+ 5^3+ ...+97^3+ 99^3
C= 1^3+ 2^3- 3^3+ 4^3- 5^3+ 6^3- ...+98^3- 99^3+ 100^3
(1/3+1/3^2+1/3^3+1/3^4).3^5+(1/3^5+1/3^6+1/3^7+1/3^8).3^9+.....+(1/3^97+1/3^98+1/3^99+1/3^100).3^101
Tính:
\(A=\frac{3^3+1^3}{2^3-1^3}+\frac{5^3+2^3}{3^3-2^3}+..............+\frac{4013^3+2006^3}{2007^3-2006^3}\)
1^3+2^3+3^3+3^4+3^5+3^6+3^7+3^8+3^9
Tính giá trị của biểu thức:
A=(1/3+1,3^2+1/3^3+1/3^4).3^5+(1/3^5+1/3^6+1/3^7+1/3^8).3^9+...+(1/3^97+1/3^98+1/3^99+1/3^100).3^101
rút gọn 1^3+2^3+3^3+...+n^3
rút gọn 1^3+3^3+5^3+...+(2n+1)^3
Cho b^2 = ac ; c^2 = bd với b, c, d ≠ 0; b+c ≠ 0; b^3+c^3≠ d^3 3. Chứng minh rằng:
a) \(\dfrac{a^3+b^3-c^3}{b^3+c^3-d^3}=\left(\dfrac{a+b-c}{b+c-d}\right)^3\)
b) \(\dfrac{a^3+b^3+c^3}{b^3+c^3+d^3}=\dfrac{a}{d}\)
A=1+3+3^2+3^3+3^4+3^5+3^6+3^7 chung minh A=(3^8-1):2