Math問題

lkhg · 發表於 12/9/2009 01:07 PM

9^(n+1)-3^(2n)
__________________
3^(2n-1)+3^(2n+1)

呢題我計左勁耐都計唔到
答案係12/5 但我計黎計去都計唔到似樣d既答案
求助

sapphire · 發表於 12/9/2009 01:30 PM

本帖最後由 sapphire 於 12/9/2009 01:31 PM 編輯

[9^(n+1) - 3^(2n)] / [3^(2n-1) - 3^(2n+1)]
= [3^(2n+2) - 3^(2n)] / [3^(2n-1) + 3^(2n+1)]
= [9 * 3^(2n) - 3^(2n)] / [3^(2n-1) + 9 * 3^(2n - 1)]
= [3^(2n) (9 - 1)] / [3^(2n-1) (1 + 9)]
= 3 * (8 / 10)
= 12 / 5

lkhg · 發表於 12/9/2009 01:47 PM

[9^(n+1) - 3^(2n)] / [3^(2n-1) - 3^(2n+1)]
= [3^(2n+2) - 3^(2n)] /[3^(2n-1) + 3^(2n+1)]
= [9 * 3^(2n) - 3^(2n)] /[3^(2n-1) + 9 * 3^(2n - 1)]
= [3^(2n) (9 - 1)] / [3^(2n-1) (1 + 9)]
= 3 * (8 / 10)
= 12 / 5
sapphire 發表於 12/9/2009 01:30 PM

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紅字唔明...?

sapphire · 發表於 12/9/2009 01:54 PM

x^a * x^b = x^(a+b)

similarly,
3^(2n+1)
= 3^(2n-1+2)
= 3^2 * 3^(2n-1)
= 9 * 3^(2n-1)

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