问题如下图：
请问为什么现值时间点P0时刻的dividend不考虑？换句话说 公式里T为什么不是从0开始?
选项：
A. 
B. 
C. 
解释：
NO.PZ2018103102000090 问题如下 Matt hevaluating the value of Company M of $20 per share using the two stage mol. The company hpaia vinof $1 per share for the most recent year. The relative highly growth rate is 5% over the first three years. The requirerate of return is 8% What`s the long-term growth rate? A.2.00% B.2.51% C.3.00% B is correct.考点Two Stage Mol解析根据公式: 20=∑t=131×(1+0.05)t(1+0.08)t+1×1.053×(1+gL)(0.08−gL)(1+0.08)320=\sum_{t=1}^3\frac{1\times\left(1+0.05\right)^t}{\left(1+0.08\right)^t}+\frac{1\times1.05^3\times\left(1+g_L\right)}{\left(0.08-g_L\right)\left(1+0.08\right)^3}20=∑t=13(1+0.08)t1×(1+0.05)t+(0.08−gL)(1+0.08)31×1.053×(1+gL)求得gL = 2.51% 如题
NO.PZ2018103102000090 问题如下 Matt hevaluating the value of Company M of $20 per share using the two stage mol. The company hpaia vinof $1 per share for the most recent year. The relative highly growth rate is 5% over the first three years. The requirerate of return is 8% What`s the long-term growth rate? A.2.00% B.2.51% C.3.00% B is correct.考点Two Stage Mol解析根据公式: 20=∑t=131×(1+0.05)t(1+0.08)t+1×1.053×(1+gL)(0.08−gL)(1+0.08)320=\sum_{t=1}^3\frac{1\times\left(1+0.05\right)^t}{\left(1+0.08\right)^t}+\frac{1\times1.05^3\times\left(1+g_L\right)}{\left(0.08-g_L\right)\left(1+0.08\right)^3}20=∑t=13(1+0.08)t1×(1+0.05)t+(0.08−gL)(1+0.08)31×1.053×(1+gL)求得gL = 2.51% =1，=1.05，=1.1025，=1.1576，V3=1.1576*（1+g）/(0.08-g)=20，反算出g=2.09%
NO.PZ2018103102000090问题如下Matt hevaluating the value of Company M of $20 per share using the two stage mol. The company hpaia vinof $1 per share for the most recent year. The relative highly growth rate is 5% over the first three years. The requirerate of return is 8% What`s the long-term growth rate?A.2.00%B.2.51%C.3.00%B is correct.考点Two Stage Mol解析根据公式: 20=∑t=131×(1+0.05)t(1+0.08)t+1×1.053×(1+gL)(0.08−gL)(1+0.08)320=\sum_{t=1}^3\frac{1\times\left(1+0.05\right)^t}{\left(1+0.08\right)^t}+\frac{1\times1.05^3\times\left(1+g_L\right)}{\left(0.08-g_L\right)\left(1+0.08\right)^3}20=∑t=13(1+0.08)t1×(1+0.05)t+(0.08−gL)(1+0.08)31×1.053×(1+gL)求得gL = 2.51%这句话我的理解是 v1=1 v2=1.05 v3=1.05^2 这明显和答案不一致 感觉很不太能够理解
2.51% 3.00% B is correct. 考点Two Stage Mol 解析根据公式: 20=∑t=131×(1+0.05)t(1+0.08)t+1×1.053×(1+gL)(0.08−gL)(1+0.08)320=\sum_{t=1}^3\frac{1\times\left(1+0.05\right)^t}{\left(1+0.08\right)^t}+\frac{1\times1.05^3\times\left(1+g_L\right)}{\left(0.08-g_L\right)\left(1+0.08\right)^3}20=∑t=13(1+0.08)t1×(1+0.05)t+(0.08−gL)(1+0.08)31×1.053×(1+gL) 求得gL = 2.51%这个题为什么不用H mol？
NO.PZ2018103102000090 2.51% 3.00% B is correct. 考点Two Stage Mol 解析根据公式: 20=∑t=131×(1+0.05)t(1+0.08)t+1×1.053×(1+gL)(0.08−gL)(1+0.08)320=\sum_{t=1}^3\frac{1\times\left(1+0.05\right)^t}{\left(1+0.08\right)^t}+\frac{1\times1.05^3\times\left(1+g_L\right)}{\left(0.08-g_L\right)\left(1+0.08\right)^3}20=∑t=13(1+0.08)t1×(1+0.05)t+(0.08−gL)(1+0.08)31×1.053×(1+gL) 求得gL = 2.51%请问这道题有用计算器快一点的方法吗？还是只能列出式子之后一步一步算哎
