Упр.870 ГДЗ Мерзляк 9 класс (Алгебра)

1) b1 = 10, q = 3, n = 4;
2) b1 = -4, q = -1, n = 10;
3) b1 = 0,6, q = 2, n = 5;
4) b1 = 4,5, q = 1/3, n = 8;
5) b1 = -9, q = корень(3), n = 6;
6) b1 = 8, q = -1/2, n = 4.

1. $$S_4=\frac{b_1(q^4-1)}{q-1}=\frac{10(3^4-1)}{3-1}=\frac{10(81-1)}{2}=10\cdot 40=400.$$

   Ответ: $$400.$$

2. $$S_{10}=\frac{b_1(1-q^{10})}{1-q}=\frac{-4(1-(-1)^{10})}{1-(-1)}=\frac{-4(1-1)}{2}=0.$$

   Ответ: $$0.$$

3. $$S_5=\frac{b_1(q^5-1)}{q-1}=\frac{0{,}6(2^5-1)}{2-1}=0{,}6\cdot 31=18{,}6.$$

   Ответ: $$18{,}6.$$

4. $$S_8=\frac{b_1(1-q^8)}{1-q}=\frac{4{,}5\left(1-\left(\frac13\right)^8\right)}{1-\frac13}.$$

   $$S_8=\frac{4{,}5\left(1-\frac1{6561}\right)}{\frac23} =4{,}5\cdot \frac{6560}{6561}\cdot \frac32 =\frac{14760}{2187} =\frac{1640}{243} =6\frac{182}{243}.$$

   Ответ: $$6\frac{182}{243}.$$

5. $$S_6=\frac{b_1(q^6-1)}{q-1}=\frac{-9\left((\sqrt3)^6-1\right)}{\sqrt3-1}=\frac{-9(27-1)}{\sqrt3-1}.$$

   $$S_6=\frac{-234}{\sqrt3-1}\cdot \frac{\sqrt3+1}{\sqrt3+1} =\frac{-234(\sqrt3+1)}{3-1} =-117(\sqrt3+1).$$

   Ответ: $$-117(\sqrt3+1).$$

6. $$S_4=\frac{b_1(1-q^4)}{1-q}=\frac{8\left(1-\left(-\frac12\right)^4\right)}{1-(-\frac12)}=\frac{8\left(1-\frac1{16}\right)}{1+\frac12}.$$

   $$S_4=8\cdot \frac{15}{16}\cdot \frac{2}{3}=5.$$

   Ответ: $$5.$$