Найдите сумму бесконечной геометрической прогрессии, проверив предварительно, что ее знаменатель q удовлетворяет условию |q|<1
а)36;12;4...
г)√2;1;1/√2
Срочно!!!
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а)
![36; 12; 4;....\
b_1=36;\
b_2=12;
b_3=4;\
q=frac{b_{n+1}}{b_{n}}=frac{b_2}{b_1}=frac{12}{36}=frac{b_3}{b_2}=frac{4}{12}=frac13;\
|q|=|frac13|<1;\
S=frac{b_1}{1-q}=frac{36}{1-frac13}=frac{36}{frac23}=18cdot3=54.](https://tex.z-dn.net/?f=36%3B+12%3B+4%3B....%5C%0Ab_1%3D36%3B%5C%0A+b_2%3D12%3B%0A+b_3%3D4%3B%5C%0Aq%3Dfrac%7Bb_%7Bn%2B1%7D%7D%7Bb_%7Bn%7D%7D%3Dfrac%7Bb_2%7D%7Bb_1%7D%3Dfrac%7B12%7D%7B36%7D%3Dfrac%7Bb_3%7D%7Bb_2%7D%3Dfrac%7B4%7D%7B12%7D%3Dfrac13%3B%5C%0A%7Cq%7C%3D%7Cfrac13%7C%26lt%3B1%3B%5C%0AS%3Dfrac%7Bb_1%7D%7B1-q%7D%3Dfrac%7B36%7D%7B1-frac13%7D%3Dfrac%7B36%7D%7Bfrac23%7D%3D18cdot3%3D54. "36; 12; 4;....\
b_1=36;\
b_2=12;
b_3=4;\
q=frac{b_{n+1}}{b_{n}}=frac{b_2}{b_1}=frac{12}{36}=frac{b_3}{b_2}=frac{4}{12}=frac13;\
|q|=|frac13|<1;\
S=frac{b_1}{1-q}=frac{36}{1-frac13}=frac{36}{frac23}=18cdot3=54.")
г)
![sqrt2; 1; frac{1}{sqrt2};....\
b_1=sqrt2;\
b_2=1;\
b_3=frac{1}{sqrt2};\
q=frac{b_{n+1}}{b_{n}}=frac{b_2}{b_1}=frac{1}{sqrt2}=frac{b_3}{b_2}=frac{frac{1}{sqrt2}}{1}=frac{1}{sqrt2};\
|q|=|frac{1}{sqrt2}|<1;\
S=frac{b_1}{1-q}=frac{sqrt2}{1-frac{1}{sqrt2}}=frac{sqrt2}{frac{sqrt2-1}{sqrt2}}=frac{sqrt2cdotsqrt2}{sqrt2-1}=frac{2}{sqrt2-1}=frac{2(sqrt2+1)}{(sqrt2)^2-1^2}=\
=frac{2sqrt2+2}{2-1}=frac{2sqrt2+2}{1}=2sqrt2+2.](https://tex.z-dn.net/?f=sqrt2%3B+1%3B+frac%7B1%7D%7Bsqrt2%7D%3B....%5C%0Ab_1%3Dsqrt2%3B%5C%0A+b_2%3D1%3B%5C%0A+b_3%3Dfrac%7B1%7D%7Bsqrt2%7D%3B%5C%0Aq%3Dfrac%7Bb_%7Bn%2B1%7D%7D%7Bb_%7Bn%7D%7D%3Dfrac%7Bb_2%7D%7Bb_1%7D%3Dfrac%7B1%7D%7Bsqrt2%7D%3Dfrac%7Bb_3%7D%7Bb_2%7D%3Dfrac%7Bfrac%7B1%7D%7Bsqrt2%7D%7D%7B1%7D%3Dfrac%7B1%7D%7Bsqrt2%7D%3B%5C%0A%7Cq%7C%3D%7Cfrac%7B1%7D%7Bsqrt2%7D%7C%26lt%3B1%3B%5C%0AS%3Dfrac%7Bb_1%7D%7B1-q%7D%3Dfrac%7Bsqrt2%7D%7B1-frac%7B1%7D%7Bsqrt2%7D%7D%3Dfrac%7Bsqrt2%7D%7Bfrac%7Bsqrt2-1%7D%7Bsqrt2%7D%7D%3Dfrac%7Bsqrt2cdotsqrt2%7D%7Bsqrt2-1%7D%3Dfrac%7B2%7D%7Bsqrt2-1%7D%3Dfrac%7B2%28sqrt2%2B1%29%7D%7B%28sqrt2%29%5E2-1%5E2%7D%3D%5C%0A%3Dfrac%7B2sqrt2%2B2%7D%7B2-1%7D%3Dfrac%7B2sqrt2%2B2%7D%7B1%7D%3D2sqrt2%2B2. "sqrt2; 1; frac{1}{sqrt2};....\
b_1=sqrt2;\
b_2=1;\
b_3=frac{1}{sqrt2};\
q=frac{b_{n+1}}{b_{n}}=frac{b_2}{b_1}=frac{1}{sqrt2}=frac{b_3}{b_2}=frac{frac{1}{sqrt2}}{1}=frac{1}{sqrt2};\
|q|=|frac{1}{sqrt2}|<1;\
S=frac{b_1}{1-q}=frac{sqrt2}{1-frac{1}{sqrt2}}=frac{sqrt2}{frac{sqrt2-1}{sqrt2}}=frac{sqrt2cdotsqrt2}{sqrt2-1}=frac{2}{sqrt2-1}=frac{2(sqrt2+1)}{(sqrt2)^2-1^2}=\
=frac{2sqrt2+2}{2-1}=frac{2sqrt2+2}{1}=2sqrt2+2.")
г)
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