Prove , by mathematical induction , that
1x3+2x3^2+3x3^3+......+n(3^n)=3/4(3^n(2n-1)+1)for all positive integer n .
SP: You only need to do the part about......
Assume S(k) is true ...and below thanks
question about M.I
2007-10-19 5:34 am
回答 (2)
2007-10-19 5:44 am
✔ 最佳答案
As follows~~~圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/ScreenHunter_04Oct182143.jpg?t=1192715058
參考: Myself~~~
2007-10-19 5:42 am
Assume S(k) is true,
i.e. 1x3+2x3^2+3x3^3+......+k(3^k)=3/4(3^k(2k-1)+1)
When n=k+1,
1x3+2x3^2+3x3^3+......+k(3^k) + (k+1) * 3^ (k+1)
= 3/4(3^k(2k-1)+1) + (k+1) * 3^ (k+1)
= 3/4[3^k(2k-1)+1 + (k+1) * 3^k * 3 * 4/3 ]
= 3/4[3^k(2k-1)+1 + 4(k+1) * 3^k ]
= 3/4[3^k(2k-1+ 4(k+1)) + 1 ]
= 3/4(3^k(2(k+1)-1)+1)
so S(k+1) is also true
2007-10-18 21:45:41 補充：
sorry, some typing mistake for the last lineit should be ---= 3/4[3^k(6k 3) 1 ]= 3/4[3^(k 1) (2k 1) 1 ]= 3/4[3^(k 1) (2(k 1)-1) 1]
i.e. 1x3+2x3^2+3x3^3+......+k(3^k)=3/4(3^k(2k-1)+1)
When n=k+1,
1x3+2x3^2+3x3^3+......+k(3^k) + (k+1) * 3^ (k+1)
= 3/4(3^k(2k-1)+1) + (k+1) * 3^ (k+1)
= 3/4[3^k(2k-1)+1 + (k+1) * 3^k * 3 * 4/3 ]
= 3/4[3^k(2k-1)+1 + 4(k+1) * 3^k ]
= 3/4[3^k(2k-1+ 4(k+1)) + 1 ]
= 3/4(3^k(2(k+1)-1)+1)
so S(k+1) is also true
2007-10-18 21:45:41 補充：
sorry, some typing mistake for the last lineit should be ---= 3/4[3^k(6k 3) 1 ]= 3/4[3^(k 1) (2k 1) 1 ]= 3/4[3^(k 1) (2(k 1)-1) 1]
收錄日期: 2021-04-13 20:03:40
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