Prove , by mathematical induction , that
1^2-2^2+3^2-4^2+......+(2n-1)^2-(2n)^2 = -n(2n+1)for all positive integer n .
Hence , find the value of
1x2-2x3+3x4-....+49x50
朋友, 你們有什麼方法去計M.I的呢??WHY我次次做到n=k+1 的時候都計唔到ge
It too difficult !!!!!! teach me how to do . (little bit afraid of CE....) 死啦!!!
M.I question
2007-10-19 6:37 am
回答 (2)
2007-10-19 6:51 am
✔ 最佳答案
As follows~~~圖片參考：http://i182.photobucket.com/albums/x4/A_Hepburn_1990/MI30.jpg?t=1192719072
參考: Myself~~~
2007-10-19 7:14 am
Assume P(n) be the statement "1^2-2^2+3^2-4^2+......+(2n-1)^2-(2n)^2 = -n(2n+1)"
When n=1, L.H.S.= [2(1)-1]^2-[2(1)]^2 = -3
. R.H.S.= -(1)[2(1)+1] = -3 = L.H.S.
Hence, P(1) is true
Assume P(k) is true for some positive integers k
i.e. 1^2-2^2+3^2-4^2+......+(2k-1)^2-(2k)^2 = -k(2k+1)
When n=k+1, L.H.S. = 1^2-2^2+3^2-4^2+......+(2k-1)^2-(2k)^2+[2(k+1)-1]^2-[2(k+1)]^2
. = -k(2k+1) + [2(k+1)-1]^2 - [2(k+1)]^2 (by assumption)
. = -k(2k+1) + (2k+1)^2 - 4(k+1)^2
. = (2k+1)[-k+(2k+1)] - 4(k+1)^2
. = (2k+1)(k+1)-4(k+1)^2
. = (k+1)[(2k+1)-4(k+1)]
. = (k+1)(2k+1-4k-4)
. = (k+1)(-2k-3)
. = -(k+1)[2(k+1)+1]
Thus, P(k+1) is true
By the mathematical induction, P(n) is true for all positive integers n.
When n=1, L.H.S.= [2(1)-1]^2-[2(1)]^2 = -3
. R.H.S.= -(1)[2(1)+1] = -3 = L.H.S.
Hence, P(1) is true
Assume P(k) is true for some positive integers k
i.e. 1^2-2^2+3^2-4^2+......+(2k-1)^2-(2k)^2 = -k(2k+1)
When n=k+1, L.H.S. = 1^2-2^2+3^2-4^2+......+(2k-1)^2-(2k)^2+[2(k+1)-1]^2-[2(k+1)]^2
. = -k(2k+1) + [2(k+1)-1]^2 - [2(k+1)]^2 (by assumption)
. = -k(2k+1) + (2k+1)^2 - 4(k+1)^2
. = (2k+1)[-k+(2k+1)] - 4(k+1)^2
. = (2k+1)(k+1)-4(k+1)^2
. = (k+1)[(2k+1)-4(k+1)]
. = (k+1)(2k+1-4k-4)
. = (k+1)(-2k-3)
. = -(k+1)[2(k+1)+1]
Thus, P(k+1) is true
By the mathematical induction, P(n) is true for all positive integers n.
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