2017-06-29 10:18:38ohe09nz81r
a.maths (mi)
標題:
a.maths (mi)
發問:
(a)prove,by mathematical induction,that 1*2+3*4+5*6+.....+(2n-1)(2n)={n(n+1)(4n-1)}/3 for all positive integers n . (b) hence,find the values of (1) 1*2+3*4+5*6+...+99*100 and (2) 51*52+53*54+55*56+....+99*100
最佳解答:
(a) let P(n) be the proposition forP(1),L.H.S.=1x2=2 R.H.S.=1x2x3/3=2 so P(1) is true assume that P(k) is true for all +ve intergers i.e.1*2+3*4+5*6+.....+(2k-1)(2k)={k(k+1)(4k-1)}/3 forP(k+1),1*2+3*4+5*6+.....+(2k-1)(2k)+(2k+1)(2(k+1)) =k(k+1)(4k-1)/3+2(2k+1)(k+1) =(k+1)(k(4k-1)+6(2k+1))/3 =(k+1)(4k^2+11k+6)/3 =(k+1)(k+2)(4k+3)/3 =(k+1)((k+1)+1)(4(k+1)-1)/3 so if P(k) is true,P(k+1) is also true so by M.i.P(n) is true for all positive integers n b(1)1*2+3*4+5*6+...+99*100 =1*2+3*4+5*6+...+(2x50-1)(2x50) =50(50+1)(4(50)-1)/3 =169150 (2)51*52+53*54+55*56+....+99*100=1*2+3*4+5*6+...+99*100 -(1*2+3*4+5*6+...+49x50) =169150-25(25+1)(4(25)-1)/3 =147700
其他解答:
為左方便睇，d乘號我用括號寫。 a) For n=1, LHS = 2 RHS = [1(1+1)(4-1)]/3 = 2 = LHS ∴ the statement is true for n=1 Assume 1(2)+3(4)+5(6)+...+(2k-1)(2k) = [k(k+1)(4k-1)]/3 1(2)+3(4)+5(6)+...+(2k-1)(2k) + 2(2k+1)(k+1) = [k(k+1)(4k-1)]/3 + 2(2k+1)(k+1) = {(k+1)[k(4k-1) + 6(2k+1)]} / 3 = [(k+1)(4k^2 + 11k + 6)] / 3 = [(k+1)(k+2)(4k+3)] / 3 ∴ the statement is true for all positive integers n bi) 1*2+3*4+5*6+...+99*100 = 50(51)(199)/3 = 169150 bii) 51*52+53*54+55*56+....+99*100 = 1*2+3*4+...+99*100 - (1*2+3*4+...+49*50) = 50(51)(199)/3 - 25(26)(99)/3 = 169150 - 21450 = 147700|||||Let S(n) be the statement "1*2+3*4+5*6+.....+(2n-1)(2n)={n(n+1)(4n-1)}/3" for all positive integers n n=1 LHS=1=RHS therefore S(n)is ture assume that S(k) is ture ie. 1*2+3*4+5*6+.....+(2k-1)(2k)={k(k+1)(4k-1)}/3 we want to prove that S9K+1) also ture ie. 1*2+3*4+5*6+.....+(2k+1)(2k+2)={(k+1)(k+2)(4k+3)}/3 for n=k+1 LHS= 1*2+3*4+5*6+.....+(2k-1)(2k)+(2k+1)(2k+2) ={k(k+1)(4k-1)}/3+(2k+1)(2k+2).....................by assumption = 1/3( k(k+1)(4k-1)+3(2k+1)(2k+2)) =1/3(4k^3+15k^2+17k+6) =1/3(k+1)(4k^2+11k+6) =1/3(k+1)(k+2)(4k+3) =RHS therefore S(k+1)is ture therefore by MI, all S(n) are ture for all positive integers n
a.maths (mi)
發問:
(a)prove,by mathematical induction,that 1*2+3*4+5*6+.....+(2n-1)(2n)={n(n+1)(4n-1)}/3 for all positive integers n . (b) hence,find the values of (1) 1*2+3*4+5*6+...+99*100 and (2) 51*52+53*54+55*56+....+99*100
最佳解答:
(a) let P(n) be the proposition forP(1),L.H.S.=1x2=2 R.H.S.=1x2x3/3=2 so P(1) is true assume that P(k) is true for all +ve intergers i.e.1*2+3*4+5*6+.....+(2k-1)(2k)={k(k+1)(4k-1)}/3 forP(k+1),1*2+3*4+5*6+.....+(2k-1)(2k)+(2k+1)(2(k+1)) =k(k+1)(4k-1)/3+2(2k+1)(k+1) =(k+1)(k(4k-1)+6(2k+1))/3 =(k+1)(4k^2+11k+6)/3 =(k+1)(k+2)(4k+3)/3 =(k+1)((k+1)+1)(4(k+1)-1)/3 so if P(k) is true,P(k+1) is also true so by M.i.P(n) is true for all positive integers n b(1)1*2+3*4+5*6+...+99*100 =1*2+3*4+5*6+...+(2x50-1)(2x50) =50(50+1)(4(50)-1)/3 =169150 (2)51*52+53*54+55*56+....+99*100=1*2+3*4+5*6+...+99*100 -(1*2+3*4+5*6+...+49x50) =169150-25(25+1)(4(25)-1)/3 =147700
其他解答:
為左方便睇，d乘號我用括號寫。 a) For n=1, LHS = 2 RHS = [1(1+1)(4-1)]/3 = 2 = LHS ∴ the statement is true for n=1 Assume 1(2)+3(4)+5(6)+...+(2k-1)(2k) = [k(k+1)(4k-1)]/3 1(2)+3(4)+5(6)+...+(2k-1)(2k) + 2(2k+1)(k+1) = [k(k+1)(4k-1)]/3 + 2(2k+1)(k+1) = {(k+1)[k(4k-1) + 6(2k+1)]} / 3 = [(k+1)(4k^2 + 11k + 6)] / 3 = [(k+1)(k+2)(4k+3)] / 3 ∴ the statement is true for all positive integers n bi) 1*2+3*4+5*6+...+99*100 = 50(51)(199)/3 = 169150 bii) 51*52+53*54+55*56+....+99*100 = 1*2+3*4+...+99*100 - (1*2+3*4+...+49*50) = 50(51)(199)/3 - 25(26)(99)/3 = 169150 - 21450 = 147700|||||Let S(n) be the statement "1*2+3*4+5*6+.....+(2n-1)(2n)={n(n+1)(4n-1)}/3" for all positive integers n n=1 LHS=1=RHS therefore S(n)is ture assume that S(k) is ture ie. 1*2+3*4+5*6+.....+(2k-1)(2k)={k(k+1)(4k-1)}/3 we want to prove that S9K+1) also ture ie. 1*2+3*4+5*6+.....+(2k+1)(2k+2)={(k+1)(k+2)(4k+3)}/3 for n=k+1 LHS= 1*2+3*4+5*6+.....+(2k-1)(2k)+(2k+1)(2k+2) ={k(k+1)(4k-1)}/3+(2k+1)(2k+2).....................by assumption = 1/3( k(k+1)(4k-1)+3(2k+1)(2k+2)) =1/3(4k^3+15k^2+17k+6) =1/3(k+1)(4k^2+11k+6) =1/3(k+1)(k+2)(4k+3) =RHS therefore S(k+1)is ture therefore by MI, all S(n) are ture for all positive integers n
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