obliczyć granice ciągów

[olciaa]

\(a) a_n= \frac{2^{n+1}-7 \cdot 3^{2n+1}}{1-2 \cdot 81^{n+1}}\)
\(b) a_n= \frac{5 \cdot 4^{n+2}+23 \cdot 7^{2n}}{49^{n+1}-3^{2n+2}}\)
\(c) a_n= \frac{ \sqrt{3} \cdot 3^{2n+2}-4^{n+2} }{2 \cdot 5^{n+1}-5 \cdot 9^{n+1}}\)
\(d) a_n= \sqrt{3 \cdot 4^{n+1}+6 \cdot 7^{2n+1}}\) <----pierwiastek ma być n-tego stopnia

odp:
a) 0
b) \(\frac{23}{49}\)
c) \(- \frac{ \sqrt{3} }{5}\)
d) 49

[irena]

a)
\(a_n=\frac{2\cdot2^n-21\cdot9^n}{1-2\cdot81\cdot81^n}=\frac{2\cdot2^n-21\cdot9^n}{1-162\cdot81^n}=\frac{2\cdot(\frac{2}{81})^n-21\cdot(\frac{1}{9})^n}{(\frac{1}{81})^n-162}\to\frac{0-0}{0-162}=\frac{0}{-162}=0\)

[irena]

\(b_n=\frac{5\cdot16\cdot4^n+23\cdot49^n}{49\cdot49^n-9\cdot9^n}=\frac{80\cdot4^n+23\cdot49^n}{49\cdot49^n-9\cdot9^n}=\frac{80\cdot(\frac{4}{49})^n+23}{49-9\cdot(\frac{9}{49})^n}\to\frac{0+23}{49-0}=\frac{23}{49}\)

[irena]

c)
\(c_n=\frac{9\sqrt{3}\cdot9^n-16\cdot4^n}{10\cdot5^n-45\cdot9^n}=\frac{9\sqrt{3}-16\cdot(\frac{4}{9})^n}{10\cdot(\frac{5}{9})^n-45}\to\frac{9\sqrt{3}-16\cdot0}{10\cdot0-45}=\frac{9\sqrt{3}}{-45}=-\frac{\sqrt{3}}{5}\)

[irena]

d)
\(d_n=\sqrt[n]{3\cdot4\cdot4^n+6\cdot7\cdot49^n}=\sqrt[n]{12\cdot4^n+42\cdot49^n}=\sqrt[n]{49^n(12\cdot(\frac{4}{49})^n+42)}=\\=49\cdot\sqrt[n]{12\cdot(\frac{4}{49})^n+42}\to49\cdot1=49\)