نویسنده : امیر انصاری

1. $$
   t_1=5, r=3, t_n=135
   $$
   
2. $$
   t_1=-2, r=-3, t_n=-1458
   $$
   
3. $$
   t_1=\frac{1}{3}, r=\frac{1}{2}, t_n=\frac{1}{48}
   $$
   
4. $$
   t_1=4, r=4, t_n=4096
   $$
   
5. $$
   t_1=-\frac{1}{6}, r=2, t_n=-\frac{128}{3}
   $$
   
6. $$
   t_1=\frac{p^2}{2}, r=\frac{p}{2}, t_n=\frac{p^9}{256}
   $$
   

پاسخ

1. $$
   t_n=t_1 r^{n-1}\\
   135 = 5 (3)^{n-1}\\
   \frac{135}{5} = 3^{n-1}\\
   27=3^{n-1}\\
   3^3=3^{n-1}\\
   3=n-1\\
   4=n
   $$
   
2. $$
   t_n=t_1 r^{n-1}\\
   -1458=-2 (-3)^{n-1}\\
   \frac{-1458}{-2} = 3^{n-1}\\
   729 = 3^{n-1}\\
   3^6 = 3^{n-1}\\
   6=n-1\\
   7=n
   $$
   
3. $$
   t_n=t_1 r^{n-1}\\
   \frac{1}{48}=\frac{1}{3} \biggl( \frac{1}{2} \biggr)^{n-1}\\
   \frac{\frac{1}{48}}{\frac{1}{3}} = \biggl( \frac{1}{2} \biggr)^{n-1}\\
   \frac{1}{16} = \biggl( \frac{1}{2} \biggr)^{n-1}\\
   \biggl( \frac{1}{2} \biggr)^4 = \biggl( \frac{1}{2} \biggr)^{n-1}\\
   4=n-1\\
   5=n
   $$
   
4. $$
   t_n = t_1 r^{n-1}\\
   4096=4 (4)^{n-1}\\
   \frac{4096}{4} = 4^{n-1}\\
   1024 = 4^{n-1}\\
   4^5 = 4^{n-1}\\
   5=n-1\\
   6=n
   $$
   
5. $$
   t_n=t_1 r^{n-1}\\
   -\frac{128}{3}=-\frac{1}{6} (2)^{n-1}\\
   \frac{-\frac{128}{3}}{-\frac{1}{6}}=2^{n-1}\\
   256=2^{n-1}\\
   2^8=2^{n-1}\\
   8=n-1\\
   9=n
   $$
   
6. $$
   t_n=t_1 r^{n-1}\\
   \frac{p^9}{256} = \frac{p^2}{2} \biggl( \frac{p}{2} \biggr)^{n-1}\\
   \frac{\frac{p^9}{256}}{\frac{p^2}{2}} = \biggl( \frac{p}{2} \biggr)^{n-1}\\
   \frac{p^7}{128} = \biggl( \frac{p}{2} \biggr)^{n-1}\\
   \frac{p^7}{2^7} = \biggl( \frac{p}{2} \biggr)^{n-1}\\
   \biggl( \frac{p}{2} \biggr)^7 = \biggl( \frac{p}{2} \biggr)^{n-1}\\
   7=n-1\\
   8=n
   $$