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Đầu tiên, Tính S1=1+2+3+...+n=\(\frac{n\left(n+1\right)}{2}\)
*/ Tính S2=12+22+32+...+n2
Đặt: S2'=1.2+2.3+3.4+...+n(n+1)
=>3S2'=1.2.3+2.3.3+3.4.3+...+n(n+1).3=1.2.3+2.3.(4-1)+3.4.(5-2)+...+n(n+1)[(n+2)−(n−1)]
Nhân ra và rút gọn ta được: 3S2′=n(n+1)(n+2) => S2'=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\)
Ta lại có: S2′=1.2+2.3+3.4+...+n(n+1)=(12+22+32+...+n2)+(1+2+3+...+n)=S2+S1=S2+\(\frac{n\left(n+1\right)}{2}\)
=> S2=S2'-\(\frac{n\left(n+1\right)}{2}\)=\(\frac{n\left(n+1\right)\left(n+2\right)}{3}\) -\(\frac{n\left(n+1\right)}{2}\)=\(\frac{n\left(n+1\right)\left(2n+1\right)}{6}\)
S3=
a: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+2n^2+3n^2+6n-n-2+n^3+2\)
\(=5n^2+5n=5\left(n^2+n\right)⋮5\)
b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=6n^2+30n+n+5-6n^2+3n-10n+5\)
\(=24n+10⋮2\)
d: \(=\left(n+1\right)\left(n^2+2n\right)\)
\(=n\left(n+1\right)\left(n+2\right)⋮6\)
Đặt
\(A_k=1+2+3+....+k=\frac{k\left(k+1\right)}{2}\)
\(A_{k-1}=1+2+3+....+\left(k-1\right)=\frac{k\left(k-1\right)}{2}\)
Ta có:
\(A_k^2-A_{k-1}^2=\frac{k^2\left(k+1\right)^2}{2}-\frac{\left(k-1\right)^2k^2}{2}=\frac{k^2}{2}\left(k^2+2k+1-k^2+2k-1\right)=k^3\)
Khi đó:
\(1^3=A_1^2\)
\(2^3=A_2^2-A_1^2\)
\(...........\)
\(n^3=A_n^2-A_{n-1}^2\)
Khi đó:
\(1^3+2^3+3^3+...+n^3=A_n^3=\left[\frac{n\left(n+1\right)}{2}\right]^2\)
\(\Rightarrow\sqrt{1^3+2^3+......+n^3}=\frac{n\left(n+1\right)}{2}\)
=> ĐPCM
Cách khác:
Ta sẽ đi chứng minh \(1^3+2^3+3^3+....+n^3=\left[\frac{n\left(n+1\right)}{2}\right]^2\)
Với n=1 thì mệnh đề trên đúng
Giả sử mệnh đề trên đúng với n=k ta sẽ chứng minh mệnh đề đúng với n=k+1
Ta có:
\(A_k=1^3+2^3+3^3+.....+k^3=\left[\frac{k\left(k+1\right)}{2}\right]^2\)
Ta cần chứng minh:
\(A_{k+1}=1^3+2^3+3^3+.....+\left(k+1\right)^3=\left[\frac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)
Thật vậy !
\(A_{k+1}=1^3+2^3+3^3+.....+\left(k+1\right)^3\)
\(=\left[\frac{k\left(k+1\right)}{2}\right]^2+\left(k+1\right)^3\)
\(=\frac{k^2\left(k+1\right)^2}{4}+\left(k+1\right)^3\)
\(=\left(k+1\right)^2\left(\frac{k^2}{4}+k+1\right)\)
\(=\left[\frac{\left(k+1\right)\left(k+2\right)}{2}\right]^2\)
Theo nguyên lý quy nạp ta có điều phải chứng minh.
\(S_n=\frac{1.2.3.4...n\left(n+1\right)\left(n+2\right)...2n}{1.2.3.4...n}\)
\(=\frac{1.3...\left(2n-1\right).2.4...\left(2n-2\right)2n}{1.2.3.4...n}\)
\(=\frac{1.3...\left(2n-1\right).2^n.1.2...n}{1.2...n}\)
\(=2^n.1.3...\left(2n-1\right)⋮2n\)
a, \(A=\frac{2^{12}\cdot3^5-4^6\cdot9^2}{(2^2\cdot3)^6+8^4\cdot3^5}-\frac{5^{10}\cdot7^3-25^5\cdot49^2}{(125\cdot7)^3+5^9\cdot14^3}\)
\(A=\frac{2^{12}\cdot3^5-2^{12}\cdot3^4}{2^{12}\cdot3^6+2^{12}\cdot3^5}-\frac{5^{10}\cdot7^3-5^{10}\cdot7^4}{5^9\cdot7^3+5^9\cdot2^3\cdot7^3}\)
\(A=\frac{2^{12}\cdot3^4(3-1)}{2^{12}\cdot3^5(3+1)}-\frac{5^{10}\cdot7^3(1-7)}{5^9\cdot7^3(1+2^3)}\)
\(A=\frac{2^{12}\cdot3^4\cdot2}{2^{12}\cdot3^5\cdot4}-\frac{5^{10}\cdot7^3\cdot(-6)}{5^9\cdot7^3\cdot9}=\frac{1}{6}-\frac{-10}{3}=\frac{7}{2}\)
b,\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=(3^{n+2}+3^n)-(2^{n+2}-2^n)\)
\(=(3^n\cdot3^2+3^n)-(2^n\cdot2^2-2^n)\)
\(=3^n\cdot(3^2+1)-2^n\cdot(2^2+1)\)
\(=3^n\cdot9+1-2^n\cdot4+1\)
\(=3^n\cdot10-2^n\cdot5\)
Vì \(2\cdot5⋮10\Rightarrow2^n\cdot5⋮10\)
\(3^n\cdot10⋮10\)
Vậy : ....
a: \(\left(n^2+3n-1\right)\left(n+2\right)-n^3+2\)
\(=n^3+2n^2+3n^2+6n-n-2-n^3+2\)
\(=5n^2+5n⋮5\)
b: \(\left(6n+1\right)\left(n+5\right)-\left(3n+5\right)\left(2n-1\right)\)
\(=\left(6n^2+30n+n+5\right)-\left(6n^2-3n+10n-5\right)\)
\(=6n^2+31n+5-6n^2-7n+5\)
\(=24n+10⋮2\)
a hơi dài để làm phần b trước :
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=3^n\cdot3^2-2^n\cdot2^2+3^n-2^n\)
\(=\left(3^n\cdot3^2+3^n\right)-\left(2^n\cdot2^2+2^n\right)\)
\(=3^n\cdot\left(3^2+1\right)-2^n\cdot\left(2^2+1\right)\)
\(=3^n\cdot10-2^n\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot2\cdot5\)
\(=3^n\cdot10-2^{n-1}\cdot10\)
\(=10\cdot\left(3^n-2^{n-1}\right)⋮10\left(đpcm\right)\)
\(A=\frac{2^{12}.3^5-4^6.9^2}{\left(2^3.3\right)^6+8^4.3^5}-\frac{5^{10}.7^3-25^5.49^2}{\left(125.7\right)^3+5^9.14^3}\)
\(A=\frac{2^{12}.3^5-\left(2^2\right)^6.\left(3^2\right)^2}{\left(2^3.3\right)^6+\left(2^3\right)^4.3^5}-\frac{5^{10}.7^3-\left(5^2\right)^5.\left(7^2\right)^2}{\left(5^3.7\right)^3+5^9.\left(2.7\right)^3}\)
\(A=\frac{2^{12}.3^5-2^{12}.3^4}{2^{18}.3^6+2^{12}.3^5}-\frac{5^{10}.7^3-5^{10}.7^4}{5^9.7^3+5^9.2^3.7^3}\)
\(A=\frac{2^{12}.3^4\left(3-1\right)}{2^{12}.3^5.\left(2^6-1\right)}-\frac{5^{10}.7^3.\left(1-7\right)}{5^9.7^3\left(1+2^3\right)}\)
\(A=\frac{2}{3.\left(64-1\right)}-\frac{5.\left(-6\right)}{9}\)
\(A=\frac{2}{3.63}+\frac{30}{9}\)
Tự lm tiếp Ball nhé~
a,thay n=1 vào thì sẽ bằng 24 ko chia hết cho 10 nên đề sai
b, \(5^n\left(5^2+5^1+1\right)=5^n.31\)
\(\left(3^{n+2}-2^{n+2}+3^n-2^n\right)\)
\(=3^n.3^2-2^n.2^2+3^n-2^n\)
\(=\left(3^n.9+3^n\right)-\left(2^n.4+2^n\right)\)
\(=3^n\left(9+1\right)-2^n\left(4+1\right)\)
\(=3^n\left(9+1\right)-2^{n-1}.2\left(4+1\right)\)
\(=3^n.10-2^{n-1}.10\)
\(=10\left(3^n-2^{n-1}\right)⋮10\left(ĐPCM\right)\)