我们知道斐波那契数列为
{ F 1 = 1 F 2 = 1 F n = F n − 1 + F n − 2 n > 2 \left\{\begin{matrix} F_1=1\\ F_2=1\\ F_n=F_{n-1}+F_{n-2} &&n>2 \end{matrix}\right. ⎩⎨⎧F1=1F2=1Fn=Fn−1+Fn−2n>2
那么如何求解他的前N项和呢
我们可以通过他的通项入手，斐波那契数列的通项为
F n = 1 5 ( ( 1 + 5 2 ) n − ( 1 − 5 2 ) n ) F_{n}=\frac{1}{\sqrt{5}}\left (\left ( \frac{1+\sqrt{5}}{2} \right )^n- \left ( \frac{1-\sqrt{5}}{2} \right )^n \right ) Fn=5 1((21+5 )n−(21−5 )n)
数列的递推公式求通项可以参考数列的递推公式求通项
有了通项后我们易知
S n = ∑ i = 1 n F i = 1 5 ( ( ( 1 + 5 2 ) 1 + ( 1 + 5 2 ) 2 + ⋯ + ( 1 + 5 2 ) n ) − ( ( 1 − 5 2 ) 1 + ( 1 − 5 2 ) 2 + ⋯ ( 1 − 5 2 ) n ) ) S_{n}=\sum _{i=1}^nF_{i}=\frac{1}{\sqrt{5}}\left (\left ( \left ( \frac{1+\sqrt{5}}{2} \right )^1+\left ( \frac{1+\sqrt{5}}{2} \right )^2+\cdots +\left ( \frac{1+\sqrt{5}}{2} \right )^n \right )- \left ( \left ( \frac{1-\sqrt{5}}{2} \right )^1+\left ( \frac{1-\sqrt{5}}{2} \right )^2+\cdots \left ( \frac{1-\sqrt{5}}{2} \right )^n \right )\right ) Sn=i=1∑nFi=5 1⎝⎛⎝⎛(21+5 )1+(21+5 )2+⋯+(21+5 )n⎠⎞−⎝⎛(21−5 )1+(21−5 )2+⋯(21−5 )n⎠⎞⎠⎞
由于这里就是两个等比数列所以可以用上等比数列求和公式
S n = ∑ i = 1 n F i = 1 5 ( ( 1 + 5 2 ( 1 − ( 1 + 5 2 ) n ) 1 − 1 + 5 2 ) − ( 1 − 5 2 ( 1 − ( 1 − 5 2 ) n ) 1 − 1 − 5 2 ) ) S_{n}=\sum _{i=1}^nF_{i}=\frac{1}{\sqrt{5}}\left (\left ( \frac{\frac{1+\sqrt{5}}{2}\left ( 1-\left ( \frac{1+\sqrt{5}}{2}\right )^n \right )}{1- \frac{1+\sqrt{5}}{2}} \right )- \left ( \frac{\frac{1-\sqrt{5}}{2}\left ( 1-\left ( \frac{1-\sqrt{5}}{2}\right )^n \right )}{1- \frac{1-\sqrt{5}}{2}} \right )\right ) Sn=i=1∑nFi=5 1⎝⎛⎝⎛1−21+5 21+5 (1−(21+5 )n)⎠⎞−⎝⎛1−21−5 21−5 (1−(21−5 )n)⎠⎞⎠⎞
= 1 5 ( ( ( 1 + 5 2 ) 2 ( 1 − ( 1 + 5 2 ) n ) 1 − 5 2 1 + 5 2 ) − ( ( 1 − 5 2 ) 2 ( 1 − ( 1 − 5 2 ) n ) 1 + 5 2 1 − 5 2 ) ) =\frac{1}{\sqrt{5}}\left (\left ( \frac{\left ( \frac{1+\sqrt{5}}{2} \right )^2\left ( 1-\left ( \frac{1+\sqrt{5}}{2}\right )^n \right )}{\frac{1-\sqrt{5}}{2}\frac{1+\sqrt{5}}{2}} \right )- \left ( \frac{\left ( \frac{1-\sqrt{5}}{2} \right )^2\left ( 1-\left ( \frac{1-\sqrt{5}}{2}\right )^n \right )}{\frac{1+\sqrt{5}}{2}\frac{1-\sqrt{5}}{2}} \right )\right ) =5 1⎝⎜⎛⎝⎜⎛21−5 21+5 (21+5 )2(1−(21+5 )n)⎠⎟⎞−⎝⎜⎛21+5 21−5 (21−5 )2(1−(21−5 )n)⎠⎟⎞⎠⎟⎞
= 1 5 ( ( 1 + 5 2 ) n + 2 − ( 1 − 5 2 ) n + 2 ) − 1 5 ( ( 1 + 5 2 ) 2 − ( 1 − 5 2 ) 2 ) =\frac{1}{\sqrt{5}}\left (\left ( \frac{1+\sqrt{5}}{2} \right )^{n+2}-\left ( \frac{1-\sqrt{5}}{2} \right )^{n+2}\right )-\frac{1}{\sqrt{5}}\left (\left ( \frac{1+\sqrt{5}}{2} \right )^{2}-\left ( \frac{1-\sqrt{5}}{2} \right )^{2}\right ) =5 1⎝⎛(21+5 )n+2−(21−5 )n+2⎠⎞−5 1⎝⎛(21+5 )2−(21−5 )2⎠⎞
= F n + 2 − F 2 = F n + 2 − 1 =F_{n+2}-F_2=F_{n+2}-1 =Fn+2−F2=Fn+2−1
所以也就是说
S n = F n + 2 − 1 S_n=F_{n+2}-1 Sn=Fn+2−1
实际上大多数的数列递推公式的通项是由含有等比因子的，都可以尝试这样的方法来求解和
