A=1+1/2.(1+2)+1/3.(1+2+3)+...+1/16.(1+2+3+...+16)=?
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11 tháng 7 2021
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: \(B=\dfrac{\left(3+1\right)\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{9^{16}-1}\)
\(=\dfrac{\left(3^2-1\right)\cdot\left(3^2+1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^4-1\right)\left(3^4+1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\cdot\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^8-1\right)\left(3^8+1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}\)
\(=\dfrac{\left(3^{16}-1\right)\left(3^{16}+1\right)}{2\left(3^{32}-1\right)}=\dfrac{1}{2}\)
NT
1
8 tháng 8 2021
\(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+...+\dfrac{1}{16}\left(1+2+3+...+16\right)\\ \Rightarrow A=1+\dfrac{1}{2}\cdot\dfrac{2\cdot3}{2}+\dfrac{1}{3}\cdot\dfrac{3\cdot4}{2}+...+\dfrac{1}{16}\cdot\dfrac{16\cdot17}{2}\\ \Rightarrow A=\dfrac{2}{2}+\dfrac{3}{2}+\dfrac{4}{2}+\dfrac{5}{2}+...+\dfrac{16}{2}+\dfrac{17}{2}\\ \Rightarrow A=\dfrac{1}{2}\left(2+3+4+...+17\right)=76\)
DB
1
NM
0
ND
0
NH
2
MA
19 tháng 4 2016
ta có
A = \(1+\frac{1+2}{2}+\frac{1+2+3}{3}+\frac{1+2+3+4}{4}+......+\frac{1+2+3+\text{4 +....+16}}{16}\)
xét tổng S = 1+2+3+4+5+......+n = \(\frac{\left(n+1\right)n}{2}\) lấy \(\frac{S}{n}=\frac{\frac{\left(n+1\right)n}{2}}{n}=\frac{n+1}{2}\)
ta có
A=\(1+\frac{\frac{2\left(2+1\right)}{2}}{2}+\frac{\frac{3\left(3+1\right)}{2}}{3}+\frac{\frac{4\left(4+1\right)}{2}}{4}+\frac{\frac{5\left(5+1\right)}{2}}{5}+......+\frac{\frac{16\left(16+1\right)}{2}}{16}\)
A = \(1+\frac{1+2}{2}+\frac{1+3}{2}+\frac{1+4}{2}+\frac{1+5}{2}+......+\frac{1+16}{2}\)
A = \(1+\frac{1+2+1+3+1+\text{4+1+5+1+6+.....+1+16}}{2}\)
A = \(1+\frac{151}{2}\)
A = \(\frac{153}{2}\)
DH
1
ND
6
20 tháng 7 2015
\(A=1+\frac{1}{2}\left(1+2\right)+\frac{1}{3}\left(1+2+3\right)+...+\frac{1}{16}\left(1+2+3+...+16\right)\).
\(A=1+\frac{1}{2}.3+\frac{1}{3}.6+....+\frac{1}{16}.136\)
\(A=1+1,5+2+...+8,5\)
\(A=\frac{\left(8,5+1\right)\left[\left(8,5-1\right):0,5+1\right]}{2}=76\)
BJ
0