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標題:
F.4 A.MATH (M.I.)
發問:
a) Prove by M.I. that 1^2 + 2^2 + 3^2 +...+ n^2 = 1/6 (n) (n+1) (2n+1) for all positive integers n.b) Hence, or otherwise, find the values of i) 2^2 + 4^2 + 6^2 +...+ (2n)^2 ii) 1^2 - 2^2 + 3^2 - 4^2 +...- 40^2c) Using the result obtained in (a), or otherwise, deduce a formula for the expression ... 顯示更多 a) Prove by M.I. that 1^2 + 2^2 + 3^2 +...+ n^2 = 1/6 (n) (n+1) (2n+1) for all positive integers n. b) Hence, or otherwise, find the values of i) 2^2 + 4^2 + 6^2 +...+ (2n)^2 ii) 1^2 - 2^2 + 3^2 - 4^2 +...- 40^2 c) Using the result obtained in (a), or otherwise, deduce a formula for the expression 1?1 + 2?3 + 3?5 +...+ n(2n-1) (a) 同 (b)(i) 就做左, 但其他的唔識計, 可唔可以幫下我, 詳列步驟，拜托!!!!! 俾埋答案: (b)(i): 2/3 (n) (n+1) (2n+1) (ii): -820 (c) 1/6 (n) (n+1) (4n-1)
最佳解答:
其他解答:
ii) 1^2 - 2^2 + 3^2 - 4^2 + ... - 40^2 = 1^2 + 3^2 + ... + 39^2 - 2^2 - 4^2 - ... - 40^2 = 1^2 + 3^2 + ... + 39^2 - (2^2 + 4^2 + ... + 40^2) = 1^2 + 2^2 + 3^2 + 4^2 + ... + 40^2 - 2(2^2 + 4^2 + ... + 40^2) = (1/6) (40) (40+1) (2(40)+1) - 2(2/3) (20) (20+1) (2(20)+1) = -820 ------------------------------------------------------------------------ 1?1 + 2?3 + 3?5 +...+ n(2n-1) = summation k(2k-1) where k from 1 to n = summation (2k^2 - k) = summation 2k^2 - summation k = 2 (summation k^2) - summation k = 2(1/6) (n) (n+1) (2n+1) - n(n + 1)/2 = [ 2(2n+1) - 3 ]n(n + 1)/6 = (4n - 1)n(n + 1)/6 2007-10-29 20:53:08 補充： oh, 慢左十三秒,不過我做得都幾簡單清楚投我一票, 唔該, 老友5FAD1C75CFAE8A5F
F.4 A.MATH (M.I.)
發問:
a) Prove by M.I. that 1^2 + 2^2 + 3^2 +...+ n^2 = 1/6 (n) (n+1) (2n+1) for all positive integers n.b) Hence, or otherwise, find the values of i) 2^2 + 4^2 + 6^2 +...+ (2n)^2 ii) 1^2 - 2^2 + 3^2 - 4^2 +...- 40^2c) Using the result obtained in (a), or otherwise, deduce a formula for the expression ... 顯示更多 a) Prove by M.I. that 1^2 + 2^2 + 3^2 +...+ n^2 = 1/6 (n) (n+1) (2n+1) for all positive integers n. b) Hence, or otherwise, find the values of i) 2^2 + 4^2 + 6^2 +...+ (2n)^2 ii) 1^2 - 2^2 + 3^2 - 4^2 +...- 40^2 c) Using the result obtained in (a), or otherwise, deduce a formula for the expression 1?1 + 2?3 + 3?5 +...+ n(2n-1) (a) 同 (b)(i) 就做左, 但其他的唔識計, 可唔可以幫下我, 詳列步驟，拜托!!!!! 俾埋答案: (b)(i): 2/3 (n) (n+1) (2n+1) (ii): -820 (c) 1/6 (n) (n+1) (4n-1)
最佳解答:
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a) Let S(n) be the proposition " 12+22+32+...+n2 = 1/6(n)(n+1)(2n+1)" when n = 1, L.H.S. = 12 = 1 R.H.S. = 1/6(1)(1+1)(2+1)1 = 1 ∴S(1) is true Assume that S(k) is true, 12+22+32+...+k2=1/6(k)(k+1)(2k+1) for all positive integers n. when n=k+1 12+22+32+...+k2+(k+1)2 =1/6(k)(k+1)(2k+1)+(k +1)2 =1/6(k+1)[k(2k+1)+6k+6] =1/6(k+1)[2k2+7k+6] =1/6(k+1)(k+2)(2k+3) ∴S(k) is true By MI for all values of n, S(n) is true b) Hence,or otherwise,find the values of i) 22+42+62+...+(2n)2 =(22)(12)+(22)(22)+(22)(32)+...+(22)(n2) =(22)[(12)+(22)+(32)+...+(n2)] =4*1/6(n+1)(n+2)(2n+ 3) =2/3(n)(n+1)(2n+1) ii) 12﹣22+32﹣42+...﹣402 because (12)+(22)+(32)+...+(402) =1/6(40*41*81) =22140 (22)+(42)+(62)+...+(402) =2/3(20)(21)(41) =11480 [(12)+(22)+(32)+...+(402)]-[12﹣22+32﹣42+...﹣402] =2[(22)+(42)+(62)+...+(402)] =2*11480 =22960 12﹣22+32﹣42+...﹣402 = 22140﹣22960 = -820 c) Using the result obtain in (a),or otherwise,deduce a formula for the expression 1．1+2．3+3．5+...+n(2n﹣1) 1．1+2．3+3．5+...+n(2n﹣1) =Σi(2i﹣1) (i from 1 up to n) =Σ(2i2﹣i) =2Σi2﹣Σi =1/3ｎ(n+1)(2n+1)﹣1/2n(n+1 ) =1/6n(n+1)[2(2n+1)﹣3] =1/6n(n+1)(4n﹣1)其他解答:
ii) 1^2 - 2^2 + 3^2 - 4^2 + ... - 40^2 = 1^2 + 3^2 + ... + 39^2 - 2^2 - 4^2 - ... - 40^2 = 1^2 + 3^2 + ... + 39^2 - (2^2 + 4^2 + ... + 40^2) = 1^2 + 2^2 + 3^2 + 4^2 + ... + 40^2 - 2(2^2 + 4^2 + ... + 40^2) = (1/6) (40) (40+1) (2(40)+1) - 2(2/3) (20) (20+1) (2(20)+1) = -820 ------------------------------------------------------------------------ 1?1 + 2?3 + 3?5 +...+ n(2n-1) = summation k(2k-1) where k from 1 to n = summation (2k^2 - k) = summation 2k^2 - summation k = 2 (summation k^2) - summation k = 2(1/6) (n) (n+1) (2n+1) - n(n + 1)/2 = [ 2(2n+1) - 3 ]n(n + 1)/6 = (4n - 1)n(n + 1)/6 2007-10-29 20:53:08 補充： oh, 慢左十三秒,不過我做得都幾簡單清楚投我一票, 唔該, 老友5FAD1C75CFAE8A5F
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