f.4 a.math數多題求解

1. 1/2+2/2^2+3/2^3+...+n/2^n=2-(n+2)/2^n
2-(1+2)/2^1 = 1/2
設 1/2+2/2^2+3/2^3+...+k/2^k=2-(k+2)/2^k
考慮
1/2+2/2^2+3/2^3+...+k/2^k + (k+1)/2^(k+1)
= 2 - (k+2)/2^k + (k+1)/2^(k+1)
= 2 - [2(k+2) - (k+1) ] / 2^(k+1)
= 2 - [2k+4 - k-1) ] / 2^(k+1)
= 2 - [k+3 ] / 2^(k+1)
= 2 - [(k+1) + 2 ] / 2^(k+1)
所以 當 n=k 情況成立, n=k+1 情況成立
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2. 6+24+60+...+ n(n+1)(n +2) = (n)(n+1)(n+2)(n+ 3)/4
(1)(1+1)(1+2)(1+ 3)/4 = 6
設 6+24+60+...+ k(k+1)(k +2) = (k)(k+1)(k+2)(k+ 3)/4
考慮 6+24+60+...+ k(k+1)(k +2) + (k+1)(k +2)(k+3)
= (k)(k+1)(k+2)(k+ 3)/4 + (k+1)(k +2)(k+3)
= (k+1)(k +2)(k+3) [k + 4] / 4
所以 當 n=k 情況成立, n=k+1 情況成立
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3. -1+3-5+...+ (-1)^n(2n-1) = (-1)^n(n)
(-1)^1 (1) = -1
設 -1+3-5+...+ (-1)^k (2k-1) = (-1)^k (k)
考慮 -1+3-5+...+ (-1)^k (2k-1) + (-1)^(k+1) (2(k+1) -1)
= (-1)^k (k) + (-1)^(k+1) (2(k+1) -1)
= (-1)^k (k) + (-1)^(k+1) (2k+1)
= (-1)^(k+1) [-k + 2k+1]
= (-1)^(k+1) [k+1]
所以 當 n=k 情況成立, n=k+1 情況成立
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4. 1^2/3+2^2/15+3^3/35+...+ n^2/ [(2n-1)(2n+1) ]
= (n)(n +1)/[2(2n+1)]

(1)(1 +1)/ [2(2(1)+1) ] = 1/3
設 1^2/3+2^2/15+3^3/35+...+ k^2/ [(2k-1)(2k+1) ]
= (k)(k +1)/[2(2k+1)]
考慮 1^2/3+2^2/15+3^3/35+...+ k^2/ [(2k-1)(2k+1) ]
+ (k+1)^2/ [(2(k+1)-1)(2(k+1)+1) ]
= (k)(k +1)/[2(2k+1)] + (k+1)^2/ [(2(k+1)-1)(2(k+1)+1) ]
= (k)(k +1)/[2(2k+1)] + (k+1)^2/ [(2k+1)(2k+3) ]
= { (k +1) / [2(2k+1)] } [ k + 2(k+1)/(2k+3) ]
= { (k +1) / [2(2k+1)] } { [k (2k+3)+ 2(k+1)]/ (2k+3) }
= { (k +1) / [2(2k+1)] } { [2k^2 +5k + 2]/ (2k+3) }
= { (k +1) / [2(2k+1)] } { [(2k + 1)(k + 2)]/ (2k+3) }
= (k +1)(k + 2) / [ 2(2k+3) ]
所以 當 n=k 情況成立, n=k+1 情況成立
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5. 1/15+2/105+3/315+... +n/(2n-1)(2n+1)(2n+3 )=(n)(n+1)/(2)(2n+1) (2n+3)
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