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m2數學幾條
發問:
1.solve the following equations for 0<=x<=360a) (sin2x)(sin6x)=(sin4x)^2b) cos(2x-45)sin(2x+45)=1/42. Suppose m>0 and n>=4. In the expansion of (x-2/x)^2(1+mx)^nin ascending powers of x , the constant term is 20 and there is no x term.find the values of m and n.3 by MI prove: ... 顯示更多 1.solve the following equations for 0<=x<=360 a) (sin2x)(sin6x)=(sin4x)^2 b) cos(2x-45)sin(2x+45)=1/4 2. Suppose m>0 and n>=4. In the expansion of (x-2/x)^2(1+mx)^n in ascending powers of x , the constant term is 20 and there is no x term.find the values of m and n. 3 by MI prove: 1^2+2^2+3^2+....+n^2=n/6(n+1)(2n+1) and 1^3+2^3+3^3+....+n^3=(n^2)((n+1)^2)/4 hense, evaluate (i) 4+18+48+100+....+900 (ii) 1/2+6+45/2+56+....+950
1a sin2xsin6x=sin24x -1/2[cos4x+cos2x]=sin24x cos4x+cos2x=-2(1-cos24x) cos4x+cos2x=-2+2cos24x 2cos22x-1+cos2x=-2+2(2cos22x-1)2 8cos^4(2x)-10cos22x-cos2x+1=0 (cos2x+1)(8cos32x-8cos2x-2cosx+1)=0 cos2x=1 [The latter one cannot be further factorized] 2x=0 or 2pi or 4pi x=0 or pi pr 2pi 1b cos(2x-45)sin(2x+45)=1/4 cos(2x+45-90)sin(2x+45)=1/4 cos[90-(2x+45)]sin(2x+45)=1/4 sin(2x+45)sin(2x+45)=1/4 sin(2x+45)=1/2 or -1/2 2x+45=30 or 150 or 210 or 330 or 390 or 510 or 570 or 690 or 750 2x=115 or 175 or 285 or 345 or 465 or 525 or 645 or 705 x=115/2 or 175/2 or 285/2 or 345/2 or 465/2 or 525/2 or 645/2 or 705/2 2 (x-2/x)2(1+mx)^n =(x2-4+4/x2){1+mnx+[n(n-1)/2](mx)2+[n(n-1)(n-2)/6](mx)3......} Constant term=-4+2m2n(n-1)=20 Coeff. of x=-4mn+(2/3)m3n(n-1)(n-2)=0 ∴m3n(n-1)(n-2)=6mn ∴12m(n-2)=6mn 2(n-2)=n n=4, m=1 or -1 3. The 2 MIs can be proved easily with normal techniques (i). 4+18+48+100+......+900 =13-12+23-22+33-32+43-42+......+103-102 =13+23+33+......+103-(12+22+32+......+102) =102(112)/4-10(11)(21)/6 =3025-385 =2640 (ii). 1/2+6+45/2+56+......+950 =(1+12+45+112+......+1900)/2 ={13+[22+23]+[2(32)+33]+[3(42)+43]+......+[9(102)+103]}/2 ={[13+23+33+......+103]+[1(22)+2(32)+3(42)+......+9(102)]}/2 ={3025+[(2-1)(22)+(3-1)(32)+......(10-1)(102)]}/2 =[3025+(23+33+43+......+103)-(22+32+42+......+102)]/2 =(3025+3024+384)/2 =6433/2
其他解答:
3. Let P(n)be the statement when n=1 LHS=1 RHS=1 P(1) is true. Assume that p(k) is also true,when n=k+1 1^2+2^2+3^2+...+k^2=k/6(k+1)(2k+1) LHS=1^2+2^2+3^2+....+n^2+(k+1)^2+n/6(n+1)(2n+1) =k/6(k+1)(2k+1)+(k+1)^2 =(k+1)(k+2)(k+3)/6 =RHS P(k+1) is also true. n is true for all positive integer by mathicmataic introduction.
m2數學幾條
發問:
1.solve the following equations for 0<=x<=360a) (sin2x)(sin6x)=(sin4x)^2b) cos(2x-45)sin(2x+45)=1/42. Suppose m>0 and n>=4. In the expansion of (x-2/x)^2(1+mx)^nin ascending powers of x , the constant term is 20 and there is no x term.find the values of m and n.3 by MI prove: ... 顯示更多 1.solve the following equations for 0<=x<=360 a) (sin2x)(sin6x)=(sin4x)^2 b) cos(2x-45)sin(2x+45)=1/4 2. Suppose m>0 and n>=4. In the expansion of (x-2/x)^2(1+mx)^n in ascending powers of x , the constant term is 20 and there is no x term.find the values of m and n. 3 by MI prove: 1^2+2^2+3^2+....+n^2=n/6(n+1)(2n+1) and 1^3+2^3+3^3+....+n^3=(n^2)((n+1)^2)/4 hense, evaluate (i) 4+18+48+100+....+900 (ii) 1/2+6+45/2+56+....+950
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最佳解答:1a sin2xsin6x=sin24x -1/2[cos4x+cos2x]=sin24x cos4x+cos2x=-2(1-cos24x) cos4x+cos2x=-2+2cos24x 2cos22x-1+cos2x=-2+2(2cos22x-1)2 8cos^4(2x)-10cos22x-cos2x+1=0 (cos2x+1)(8cos32x-8cos2x-2cosx+1)=0 cos2x=1 [The latter one cannot be further factorized] 2x=0 or 2pi or 4pi x=0 or pi pr 2pi 1b cos(2x-45)sin(2x+45)=1/4 cos(2x+45-90)sin(2x+45)=1/4 cos[90-(2x+45)]sin(2x+45)=1/4 sin(2x+45)sin(2x+45)=1/4 sin(2x+45)=1/2 or -1/2 2x+45=30 or 150 or 210 or 330 or 390 or 510 or 570 or 690 or 750 2x=115 or 175 or 285 or 345 or 465 or 525 or 645 or 705 x=115/2 or 175/2 or 285/2 or 345/2 or 465/2 or 525/2 or 645/2 or 705/2 2 (x-2/x)2(1+mx)^n =(x2-4+4/x2){1+mnx+[n(n-1)/2](mx)2+[n(n-1)(n-2)/6](mx)3......} Constant term=-4+2m2n(n-1)=20 Coeff. of x=-4mn+(2/3)m3n(n-1)(n-2)=0 ∴m3n(n-1)(n-2)=6mn ∴12m(n-2)=6mn 2(n-2)=n n=4, m=1 or -1 3. The 2 MIs can be proved easily with normal techniques (i). 4+18+48+100+......+900 =13-12+23-22+33-32+43-42+......+103-102 =13+23+33+......+103-(12+22+32+......+102) =102(112)/4-10(11)(21)/6 =3025-385 =2640 (ii). 1/2+6+45/2+56+......+950 =(1+12+45+112+......+1900)/2 ={13+[22+23]+[2(32)+33]+[3(42)+43]+......+[9(102)+103]}/2 ={[13+23+33+......+103]+[1(22)+2(32)+3(42)+......+9(102)]}/2 ={3025+[(2-1)(22)+(3-1)(32)+......(10-1)(102)]}/2 =[3025+(23+33+43+......+103)-(22+32+42+......+102)]/2 =(3025+3024+384)/2 =6433/2
其他解答:
3. Let P(n)be the statement when n=1 LHS=1 RHS=1 P(1) is true. Assume that p(k) is also true,when n=k+1 1^2+2^2+3^2+...+k^2=k/6(k+1)(2k+1) LHS=1^2+2^2+3^2+....+n^2+(k+1)^2+n/6(n+1)(2n+1) =k/6(k+1)(2k+1)+(k+1)^2 =(k+1)(k+2)(k+3)/6 =RHS P(k+1) is also true. n is true for all positive integer by mathicmataic introduction.
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