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標題:
F.4 AM
發問:
1)Prove that,(1/2)+(3/2^2)+(5/2^3)+....+(2n-1/2^n)=3-(2n+3/2^n) for all positive integer n. 2)Find the value of n if nC2=55
最佳解答:
Let P ( n ) be the proposition “(1/2)+(3/22)+(5/23)+....+(2n-1/2n)=3-(2n+3/2n)”. When n = 1, L.H.S. = ( 2 – 1 ) / 21 = 1 / 2 R.H.S. = 3 – ( 2 + 3 ) / 21 = 1 / 2 = L.H.S. So P ( 1 ) is true. Assume P ( k ) is true for some positive integers k, i.e. (1/2)+(3/22)+(5/23)+....+(2k-1/2k)=3-(2k+3/2k) When n = k + 1, L.H.S. = (1/2)+(3/22)+(5/23)+....+(2k-1/2k) + (2k+2-1/2k+1) = 3-(2k+3/2k) + (2k+2-1/2k+1) = 3 –(2k+3/2k) + (2k+1/2k )( 1 / 2 ) = 3 + ( 1 / 2 )[ (2k+1)/2k – (4k+6)/2k ] = 3 + ( 1 / 2 x 2k )( 2k + 1 – 4k – 6 ) = 3 + ( 1 / 2k+1 )( -2k – 5 ) = 3 – ( 2k + 5 ) / 2k+1 R.H.S. = 3 – ( 2k + 2 + 3 ) / 2k+1 = 3 – ( 2k + 5 ) / 2k+1 = L.H.S. So P ( k + 1 ) is true. By the principle of mathematical induction, P ( n ) is true for all positive integers n. 2)nC2 = 55 n! / 2! ( n – 2 )! = 55 ( 1 / 2 )( n )( n – 1 ) = 55 n2 – n = 110 n2 – n – 110 = 0 ( n – 11 )( n + 10 ) = 0 n = 11 or n = -10 ( rejected ) Hence n = 11.
其他解答:
F.4 AM
發問:
1)Prove that,(1/2)+(3/2^2)+(5/2^3)+....+(2n-1/2^n)=3-(2n+3/2^n) for all positive integer n. 2)Find the value of n if nC2=55
最佳解答:
Let P ( n ) be the proposition “(1/2)+(3/22)+(5/23)+....+(2n-1/2n)=3-(2n+3/2n)”. When n = 1, L.H.S. = ( 2 – 1 ) / 21 = 1 / 2 R.H.S. = 3 – ( 2 + 3 ) / 21 = 1 / 2 = L.H.S. So P ( 1 ) is true. Assume P ( k ) is true for some positive integers k, i.e. (1/2)+(3/22)+(5/23)+....+(2k-1/2k)=3-(2k+3/2k) When n = k + 1, L.H.S. = (1/2)+(3/22)+(5/23)+....+(2k-1/2k) + (2k+2-1/2k+1) = 3-(2k+3/2k) + (2k+2-1/2k+1) = 3 –(2k+3/2k) + (2k+1/2k )( 1 / 2 ) = 3 + ( 1 / 2 )[ (2k+1)/2k – (4k+6)/2k ] = 3 + ( 1 / 2 x 2k )( 2k + 1 – 4k – 6 ) = 3 + ( 1 / 2k+1 )( -2k – 5 ) = 3 – ( 2k + 5 ) / 2k+1 R.H.S. = 3 – ( 2k + 2 + 3 ) / 2k+1 = 3 – ( 2k + 5 ) / 2k+1 = L.H.S. So P ( k + 1 ) is true. By the principle of mathematical induction, P ( n ) is true for all positive integers n. 2)nC2 = 55 n! / 2! ( n – 2 )! = 55 ( 1 / 2 )( n )( n – 1 ) = 55 n2 – n = 110 n2 – n – 110 = 0 ( n – 11 )( n + 10 ) = 0 n = 11 or n = -10 ( rejected ) Hence n = 11.
其他解答:
此文章來自奇摩知識+如有不便請留言告知
1) for n = 1 LHS = 1/2 RHS = 3-(2+3)/2 = 1/2 Therefore, the equation is true for n = 1 Assume the equation is true for n = k ie. (1/2) + (3/2^2) + ... + (2k-1)/2^k = 3 - (2k+3)/2^k When n = k+1 LHS = (1/2) + (3/2^2) + ... + (2k-1)/2^k + [2(k+1)-1]/2^(k+1) = 3 - (2k+3)/2^k + [2(k+1)-1]/2^(k+1) = 3 - (4k+6)/2^(k+1) + [(2k+1)]/2^(k+1) = 3 - [(4k+6) - 2k - 1]/2^(k+1) = 3 - (2k+5)/2^(k+1) = 3 - [2(k+1)+3]/2^(k+1) = RHS Therefore, the equation is true for all positive integer n according to MI 2) nC2 = n (n-1)/2 = 55 n^2 - n = 110 n^ - n - 110 = 0 (n-11)(n+10) = 0 n = 11 or n = -10 (rejected) Therefore, n = 11|||||諗左好耐先知你打緊咩 -(2n+3/2^n) 應該係 [(2n+3)/(2^n)] 記住先成除後加減 Let P(n) be the proposition '(1/2)+(3/2^2)+(5/2^3)+....+[(2n-1)/2^n]=3-[(2n+3)/2^n]' when n=1 LHS = 1/2 RHS = 3-(2+3)/2 = 1/2 so P(1) is true Assume that P(k) is true for some positive integer k i.e. (1/2)+(3/2^2)+(5/2^3)+....+ [(2k-1)/2^k]=3-[(2k+3)/2^k] When n=k+1 LHS =(1/2)+(3/2^2)+(5/2^3)+....+[(2k-1)/2^k] + [2(k+1)-1]/[2^(k+1)] =3-[(2k+3)/2^k] + (2k+1)/[2^(k+1)] =3- (4k + 6) / [(2^k) x 2] +(2k+1)/[2^(k+1)] =3- (4k + 6) / [2^(k+1)] +(2k+1)/[2^(k+1)] =3 + (-4k - 6 + 2k +1) / [2^(k+1)] =3+ (-2k-5)/ [2^(k+1)] =3-(2k+5) / [2^(k+1)] RHS =3- [2(k+1)+3]/[2^(k+1)] =3- (2k+5) / [2^(k+1)] =LHS so P(k+1) is true By the principle of mathematical induction P(n) is true for all positive integers n 2) nC2 = 55 n(n-1) / 2 = 55 n^2 - n = 110 n^2 - n -110 =0 (n-11)(n+10) = 0 n = 11 or -10 (rejected as n cannot be negative) so n =11
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