trả lời được thì **** cho
Đúng 0
Bình luận (0)
Các câu hỏi tương tự
Chứng minh 2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10+2^11+2^12+2^13+2^14+2^15+2^16+2^17+2^18+...+2^79+2^80
A=4+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10+2^11+2^12+2^13+2^14+2^15+2^16+2^17+2^18+2^19+2^20
Tính A
A=1+2+2^2 +2^3 +2^4 + 2^5 +2^6 +2^7+2^8+2^9+2^10+2^11+2^12+2^13+2^14+2^15+2^16+2^17+2^18+2^19+2^20
A=?
Tinh
1) 1^2 + 3^2 + 5^2 + ... + 97^2 + 99^2
2) 1^2 - 2^2 + 3^2 - 4^2 + ... + 99^2 - 100^2
3) 1.2^2 + 2.3^2 + 3.4^2 + ... + 98.99^2
1.A=2^2+4^2+6^2+...+100^2
2.A=1^2+2^2+3^2+...+99^2
3.A=1^2+2^2+3^2+...+100^2
2+2+2+2+2
2+2+2+2+2
2+2+2+2+0 =
Mn kb mk đi
[1+2+2^2+2^3+2^4+2^5+2^6+2^7] chia hết cho 3
[1+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10+2^11] chia hết cho 9
14 tháng 5 2022 lúc 9:48
Rút gọn các tổng sau:a) A 2 - 2^2 + 2^3 - 2^4 + ... + 2^{99} - 2^{100}b) B 1 + 2^2 + 2^4 + ... + 2^{98} + 2^{100}c) C 1 - 2^3 + 2^6 - 2^9 + ... + 2^{60} - 2^{63} + 2^{69}d) D 1 + dfrac{1}{3} + dfrac{1}{3^2} + dfrac{1}{3^4} + ... + dfrac{1}{3^{100}}e) E 1 - dfrac{1}{4} + dfrac{1}{4^2} - dfrac{1}{4^3} + ... + dfrac{1}{4^{98}} - dfrac{1}{4^{99}} + dfrac{1}{4^{100}}
Đọc tiếp
Rút gọn các tổng sau:
a) A = 2 - 2\(^2\) + 2\(^3\) - 2\(^4\) + ... + 2\(^{99}\) - 2\(^{100}\)
b) B = 1 + 2\(^2\) + 2\(^4\) + ... + 2\(^{98}\) + 2\(^{100}\)
c) C = 1 - 2\(^3\) + 2\(^6\) - 2\(^9\) + ... + 2\(^{60}\) - 2\(^{63}\) + 2\(^{69}\)
d) D = 1 + \(\dfrac{1}{3}\) + \(\dfrac{1}{3^2}\) + \(\dfrac{1}{3^4}\) + ... + \(\dfrac{1}{3^{100}}\)
e) E = 1 - \(\dfrac{1}{4}\) + \(\dfrac{1}{4^2}\) - \(\dfrac{1}{4^3}\) + ... + \(\dfrac{1}{4^{98}}\) - \(\dfrac{1}{4^{99}}\) + \(\dfrac{1}{4^{100}}\)
bài 1, so sanh
a,\(\frac{2}{2^2}+\frac{2}{3^2}+\frac{2}{4^2}+...+\frac{2}{10^2}\) và 1
b,\(\frac{2}{2^2}+\frac{2}{4^2}+\frac{2}{6^2}+\frac{8}{2^2}+\frac{2}{10^2}+\frac{2}{12^2}\)và1