有几條台灣入學ge數!!!!－hzb53jl55v的部落格

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有几條台灣入學ge數!!!!

發問:

1-設n為正整數,且(1.25)^n的整數部分3位數,則這種n共有几多個? 2-分式(x^2-x+1)/(x^2+x+1)的最大值發生在x=? 3-設n是4的倍數,則1+i+2i^2+3i^3+.........+ni^n 的和是多少? i 是開方根-1 4-1*99+2*98+3*97+.......48*52+49*51+50*50的值是什麼

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最佳解答:

1-設n為正整數,且(1.25)n的整數部分3位數,則這種n共有几多個? 即 100 < 1.25n < 1000 1.25n <1000 log 1.25n < log 1000 n log 1.25 < 3 n < 30.956 n ≦ 30 1.25n > 100 n log 1.25 > 2 n > 20.6 n ≧ 21 所以共有 30 -21 + 1 = 10個 2-分式(x2-x+1)/(x2+x+1)的最大值發生在x=? y = (x2-x+1)/(x2+x+1) dy/dx = [(x2+x+1)(2x-1) – (x2-x+1)(2x+1)] / (x2+x+1)2 若有極大值，則 dy/dx = 0 (x2+x+1)(2x-1) – (x2-x+1)(2x+1) = 0 2x3+2x2+2x-x2-x-1-2x3+2x2-2x-x2+x-1 = 0 2x2 - 2 = 0 x2 = 1 x = 1 或 x = -1 當 x = 1 y = (x2-x+1)/(x2+x+1) = (1 – 1 + 1)/(1 + 1 + 1) = 1/3 當 x = -1 y = (x2-x+1)/(x2+x+1) = (1 + 1 + 1)/(1 – 1 + 1) = 3 所以當x = -1，(x2-x+1)/(x2+x+1) 有極大值。 3-設n是4的倍數,則1+i+2i2+3i3+.........+nin 的和是多少? i 是開方根-1 因 i = i i2 = -1 i3 = -i i4 = 1 i5 = i …….. i4m = 1 i4m + 1 = i i4m + 2 = -1 i4m + 3 = -i 1+i+2i2+3i3+.........+nin = 1 + i – 2 – 3i + 4 + 5i – 6 – 7i + ……….+ n = 1 – 2 + 4 – 6 + 8 - ….. + n + i – 3i + 5i – 7i +…..- (n-1)i = 1 + (4 – 2) + (8 – 6) + ……(n – (n-2)) + (1 – 3)i + (5 – 7)i +…..+ ((n-3) – (n-1)) = 1 + 2 + 2 + 2 + …. + 2 – 2i – 2i – 2i - ….. – 2i [式中共有 (n/4) 個 2 – 2i] = 1 + n(1 – i)/2 4-1*99+2*98+3*97+... ....48*52+49*51+50*5 0的值是什麼 50*50 = 2500 49*51 = 2499 = 2500 – 1 48*52 = 2496 = 2500 - 22 47x53 = 2491 = 2500 - 33 1*99+2*98+3*97+... ....48*52+49*51+50*5 0 = (2500 – 492) + (2500 – 482) + (2500 – 472) + ….. + (2500 – 12) + 2500 = 2500*50 – (12 + 22 + 32 + ….. 492) = 2500*50 – 49*(49 + 1)(2*49 + 1)/6 = 84575

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