\(1+2^2+3^2+.......+90^2=?\)
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DB
1
DN
19 tháng 12 2022
\(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{90}}\)
\(Đặt:A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{90}}\)
Ta có:
\(A=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{90}}\)
\(\dfrac{A}{2}=\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{91}}\)
\(\dfrac{A}{2}-A=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{91}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{90}}\right)\)
\(-\dfrac{1}{2}A=\dfrac{1}{2^{91}}-\dfrac{1}{2}\)
\(A=\dfrac{\left(\dfrac{1}{2^{91}}-\dfrac{1}{2}\right)}{-\dfrac{1}{2}}\)
\(A=-2\left(\dfrac{1}{2^{91}}-\dfrac{2^{90}}{2^{91}}\right)\)
\(A=\dfrac{-2\left(1-2^{90}\right)}{2^{91}}\)
\(A=\dfrac{-2-\left[-\left(2^{91}\right)\right]}{2^{91}}\)
\(A=\dfrac{-2+2^{91}}{2^{91}}\)
\(A=-\dfrac{2}{2^{91}}+\dfrac{2^{91}}{2^{91}}\)
\(A=\dfrac{1}{2^{90}}-1\)
NT
0
DL
3
B
27 tháng 4 2022
1)( \(\sqrt{2}\) +1)^3-( \(\sqrt{2}\) -1)^3=
( \(\sqrt{2}\) +1- \(\sqrt{2}\) +1)[( \(\sqrt{2}\) +1)^2+( \(\sqrt{2}\) -1)( \(\sqrt{2}\) +1)+(\(\sqrt{2}\) -1)^2]
=2( 2+\(2\sqrt{2}\)+1+2-1+2-\(2\sqrt{2}\)+1)=2.7=14
DT
1
HT
16 tháng 12 2020
\(X=\left(a+b\right)^n=\sum\limits^n_{k=0}C^k_n.a^k.b^{n-k}\)
\(\Rightarrow\left\{{}\begin{matrix}a=2\\b=1\end{matrix}\right.\)
\(\Rightarrow A=\sum\limits^{90}_{k=2}C^k_{90}.2^k=...\)
Hoặc có thể làm như vầy: \(A=X-C^0_{90}.2^0-C^1_{90}.2=3^{90}-1-90.2=...\)
27 tháng 12 2018
\(1+2+2^2+2^3+......+2^{90}\)
\(=\left(1+2+2^2+2^3\right)+....+\left(2^{87}+2^{88}+2^{89}+2^{90}\right)\)
\(=15+...+2^{87}\cdot15\)
\(=15\left(1+...+2^{87}\right)⋮15\)