几何序列益智游戏
以下问题困扰了我几天(注意:这不是作业)。
存在两个等比数列,其和为 9。它们的第二项 (t2) 的值为 2。
- 求公比 (r)
- 求每个等比数列的第一个元素 (t1)
(1) 的答案是 2/3 和 1 /3 和 (2) 的答案分别是 3 和 6。不幸的是,我不明白这些是如何得出的。
在解决 (1) 问题时,我尝试应用代数替换来求解 r,如下所示:
t2 = t1*r; since t2 = 2 we have:
t1 = 2/r
计算收敛到极限的序列之和 (S) 的方程由下式给出:
S = t1 / (1 - r)
因此,我尝试将我的值代入将 t1 代入 S 并按如下方式求解 r:
9 = (2/r) / (1-r)
9(1-r) = 2/r
2/9 = r(1-r)
不幸的是,从这一点开始我陷入了困境。我需要消除其中一个 r,但我似乎做不到。
接下来,我想使用对序列的前两项 (S2) 求和的公式来求解 r:
S2 = (t1 (1-r^2)) / (1-r)
t1 + 2 = (t1 (1-r^2)) / (1-r)
但是将其展开,我再次遇到相同的问题(无法消除其中一个 r)。
所以我有两个问题:
- 导出 r 时我做错了什么?
- 一旦我有了其中一个值,我如何导出另一个值?
The following problem has been puzzling me for a couple of days (nb: this is not homework).
There exists two geometric sequences that sum to 9. The value of their second term (t2) is 2.
- Find the common ratio (r)
- Find the first element (t1) of each
The answers to (1) are 2/3 and 1/3 and the answers to (2) are 3 and 6 respectively. Unfortunately, I can't understand how these were derived.
In tackling (1) I've tried to apply algebraic substitution to solve for r as follows:
t2 = t1*r; since t2 = 2 we have:
t1 = 2/r
The equation for calculating the sum (S) of a sequence that converges to a limit is given by:
S = t1 / (1 - r)
So, I tried to plug my value of t1 into S and solve for r as follows:
9 = (2/r) / (1-r)
9(1-r) = 2/r
2/9 = r(1-r)
Unfortunately, from this point I get stuck. I need to eliminate one of the r's but I can't seem to be able to.
Next, I thought to solve for r using the formula that sums the first 2 terms (S2) of the sequence:
S2 = (t1 (1-r^2)) / (1-r)
t1 + 2 = (t1 (1-r^2)) / (1-r)
but expanding this out I again run into the same problem (can't eliminate one of the r's).
So I have 2 questions:
- What am I doing wrong when deriving r?
- Once I have one of its values, how I derive the other?
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你需要学会因式分解!
为了更简单,让 R = 3r
并且你的第一项是 2/(1/3) = 6,或 2/(2/3) = 3
QED!
You need to learn to factorise!
to make it easier, let R = 3r
And your first term is 2/(1/3) = 6, or 2/(2/3) = 3
QED!
将其重写为 ax2 + bx + c 并使用二次公式求解:
2/9 = r - r2< br>
r2 - r + 2/9 = 0
使用二次公式,根为:
[-1 ± √(1 - 8/9)] / 2
= (1 ± 1/3) / 2
= 1/2 ± 1/6
= 1/3 或 2/3
编辑: 哎呀,我花了太长时间才弄清楚如何写加/减和平方根。 :-P
Rewrite this as ax2 + bx + c and use the quadratic formula to solve it:
2/9 = r - r2
r2 - r + 2/9 = 0
Using the quadratic formula, the roots are:
[-1 ± √(1 - 8/9)] / 2
= (1 ± 1/3) / 2
= 1/2 ± 1/6
= 1/3 or 2/3
Edit: Aw shoot, I spent way too long figuring out how to write plus/minus and square root. :-P