Упр.172 ГДЗ Колягин Ткачёва 9 класс (Алгебра)

1) a_n=-5n+4; 2) a_n=2(n-10);
3) a_n=2·3^(n+1); 4) a_n=7·(1/2)^(n+2).

1. $$a_n=-5n+4$$

   Подставим вместо $$n$$ значения $$n+1$$, $$n+2$$ и $$n+5$$:

   $$a_{n+1}=-5(n+1)+4=-5n-5+4=-5n-1$$
   
   $$a_{n+2}=-5(n+2)+4=-5n-10+4=-5n-6$$
   
   $$a_{n+5}=-5(n+5)+4=-5n-25+4=-5n-21$$

2. $$a_n=2(n-10)$$

   $$a_{n+1}=2((n+1)-10)=2(n-9)$$
   
   $$a_{n+2}=2((n+2)-10)=2(n-8)$$
   
   $$a_{n+5}=2((n+5)-10)=2(n-5)$$

3. $$a_n=2\cdot 3^{n+1}$$

   $$a_{n+1}=2\cdot 3^{(n+1)+1}=2\cdot 3^{n+2}$$
   
   $$a_{n+2}=2\cdot 3^{(n+2)+1}=2\cdot 3^{n+3}$$
   
   $$a_{n+5}=2\cdot 3^{(n+5)+1}=2\cdot 3^{n+6}$$

4. $$a_n=7\cdot \left(\frac12\right)^{n+2}$$

   $$a_{n+1}=7\cdot \left(\frac12\right)^{(n+1)+2}=7\cdot \left(\frac12\right)^{n+3}$$
   
   $$a_{n+2}=7\cdot \left(\frac12\right)^{(n+2)+2}=7\cdot \left(\frac12\right)^{n+4}$$
   
   $$a_{n+5}=7\cdot \left(\frac12\right)^{(n+5)+2}=7\cdot \left(\frac12\right)^{n+7}$$

Ответ

1) $$a_{n+1}=-5n-1,\ a_{n+2}=-5n-6,\ a_{n+5}=-5n-21$$;
2) $$a_{n+1}=2(n-9),\ a_{n+2}=2(n-8),\ a_{n+5}=2(n-5)$$;
3) $$a_{n+1}=2\cdot 3^{n+2},\ a_{n+2}=2\cdot 3^{n+3},\ a_{n+5}=2\cdot 3^{n+6}$$;
4) $$a_{n+1}=7\cdot \left(\frac12\right)^{n+3},\ a_{n+2}=7\cdot \left(\frac12\right)^{n+4},\ a_{n+5}=7\cdot \left(\frac12\right)^{n+7}$$.