M=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+...+16) = ?
Những câu hỏi liên quan
thuc hien phep tinh :M=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+...+16)
Thực hiện tính
M=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+...+16)
Thực hiện phép tính:
M=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+...+16)
A, thực hiện phép tính
M=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+4+...+16)
B, tìm x biết
|4x+3|-|x-1|=7
A=1+1/2(1+2)+1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+4+...+16)
E=1+1/2×(1+2)+1/3×(1+2+3)+1/4×(1+2+3+4)+...+1/16×(1+2+3+4+...+16)
tính hợp lý 1 + 1/2(1+2) +1/3(1+2+3)+1/4(1+2+3+4)+...+1/16(1+2+3+4+...+16)
rút gọn biểu thức
a) A=16^8 -1/(2+1)(2^2+1)(2^4+1)(2^8+1(3^16+1)
b) B=(3+1)(3^2+1)(3^4+1)(3^8+1)(3^16+1)/9^16-1
giúp mk vs ah mk đang cần gấp ah
a) Ta có: \(A=\dfrac{16^8-1}{\left(2+1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^2-1\right)\left(2^2+1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^4-1\right)\left(2^4+1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^8-1\right)\left(2^8+1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{\left(2^{16}-1\right)\left(2^{16}+1\right)}\)
\(=\dfrac{2^{32}-1}{2^{32}-1}=1\)
b) Ta có: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)
Đúng 0
Bình luận (1)
Tính P = 1+ 1/2(1+2) + 1/3(1+2+3) + 1/4(1+2+3+4)+...+1/16(1+2+3+...+16)