標題:
幾題數學歸納法
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利用數學歸納法證明:對於所有正整數n,各命題成立 1.)5+2*5^2+3*5^3+...........+n*5^n=(4n-1)5^n+1+5/16 2.)a+ar+ar^2+.........ar^n-1=a(r^n-1)/r-1,其中r=/1 更新: 第一題打錯左 應該係5+2*5^2+3*5^3+...........+n*5^n=((4n-1)5^n+1+5)/16
最佳解答:
1.設S(n)為以下命題： 5+2*52+3*53+...........+n*5n= [(4n﹣1)5n+1+5]/16 當n = 1時， L.H.S.=1*51= 5 R.H.S.=[(4﹣1)*52+5]/16 = 80/16=5 ∴L.H.S.=R.H.S. ∴ ∴S(1)成立 假設S(k)成立， 即5+2?52+3?53+...+k?5k= [(4k﹣1)5k+1+5]/16 L.H.S. =5+2?52+3?53+...+k?5k+(k+1)?5k+1 =[(4k﹣1)5k+1+5]/16 + (k+1)?5k+1 =[(4k﹣1)5k+1+5]/16 + 16(k+1)?5k+1/16 =[(4k﹣1)5k+1+5 + 16(k+1)?5k+1]/16 =[(4k﹣1+16k+16)?5k+1+5]/16 =[(20k+15)?5k+1+5]/16 =[(4k+3)?5?5k+1+5]/16 =[(4k+3)?5k+2+5]/16 = {[4(k+1)﹣1]?5(k+1)+1+5}/16 ∴ ∴S(k)成立 根據數學歸納法的原理，對於所有正整數n，S(n)成立 2.設S(n)為以下命題： a+ar+ar2+...+arn-1 = a(rn﹣1)/r﹣1 當n = 1時， L.H.S.=ar1-1 = a(r0) = a R.H.S.= a(r1﹣1)/r﹣1 = a ∴L.H.S.=R.H.S. ∴S(1)成立 假設S(k)成立， 即a+ar+ar2+...+ark-1 = a(rk﹣1)/r﹣1 L.H.S. =a+ar+ar2+...+ark-1+ ar(k+1)-1 =[a(rk﹣1)/r﹣1]+ ar(k+1)-1 =[a(rk﹣1)/r﹣1]+ ark =[a(rk﹣1)/r﹣1]+ [ark(r﹣1)/(r﹣1)] =[a(rk﹣1)+ ark(r﹣1)]/(r﹣1) = a[rk﹣1+ rk(r﹣1)]/(r﹣1) = a(rk﹣1+ rk+1﹣rk)/(r﹣1) = a(rk+1﹣1)/(r﹣1) = R.H.S. ∴S(k)成立 根據數學歸納法的原理，對於所有正整數n，S(n)成立
其他解答:
hey, seems like you have some typing mistake for the questions or you are missing some important brackets, the case is not correct even for n =1 please check your question again~ 2007-11-01 17:14:39 補充： typing mistake for the question 1 i mean
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