
The quotient when $1 + {x^2} + {x^4} + {x^6} + .......... + {x^{34}}$ is divided by $1 + x + {x^2} + {x^3} + ........... + {x^{17}}$. is
A. ${x^{17}} - {x^{15}} + {x^{13}} - {x^{11}} + ......x$
B. ${x^{17}} + {x^{15}} + {x^{13}} + {x^{11}} + ..........x$
C. ${x^{17}} + {x^{16}} + {x^{15}} + {x^{14}} + ..........x$
D. ${x^{17}} - {x^{16}} + {x^{15}} - {x^{14}} + .......... + x$
Answer
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Hint: In order to solve this problem we need to use the formula of sum of finite geometric series $S = \dfrac{{a({r^n} - 1)}}{{r - 1}}$ . Then we need to use it for both the series given and then we need to divide the obtained sum to get the right answer.
Complete step-by-step answer:
As we know, the sum of finite geometric series $1,r,{r^2},{r^3},{r^4}..........{r^{n - 1}}$ with common ratio is given by
$a,ar,a{r^2},a{r^3},a{r^4}..........a{r^{n - 1}} = S = \dfrac{{a({r^n} - 1)}}{{r - 1}}$ = sum (where r is greater than 1)
When r is less than 1 then the formula of sum for n terms of GP will be $\dfrac{{a(1 - {r^n})}}{{1 - r}}$.
Now, $1 + {x^2} + {x^4} + {x^6} + .......... + {x^{34}}$ forms a finite geometric series with common ratio ${x^2}$, first term 1 and number of terms n = 18 (since power of x from 2 to 34 with the difference of 2 is 17 terms and the first term is 1 therefore n = 18) then sum is,
$1 + {x^2} + {x^4} + {x^6} + .......... + {x^{34}} = \dfrac{{{x^{2(18)}} - 1}}{{{x^2} - 1}}$
$1 + {x^2} + {x^4} + {x^6} + .......... + {x^{34}} = \dfrac{{{x^{36}} - 1}}{{{x^2} - 1}}$……….(1)
Now, $1 + x + {x^2} + {x^3} + ........... + {x^{17}}$ forms a finite geometric series with common ratio x, first term 1 and number of terms n = 18 (since power of x from 1 to 17 is 17 terms and first term is 1 therefore n = 18) then sum is
$1 + x + {x^2} + {x^3} + ........... + {x^{17}} = \dfrac{{{x^{18}} - 1}}{{x - 1}}$……………..(2)
Now, Dividing (1) by (2) equation
$ \Rightarrow \dfrac{{\left( {{x^{36}} - 1} \right)\left( {x - 1} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^{18}} - 1} \right)}} = \dfrac{{{x^{18}} - 1}}{{x + 1}}$
Now, we have to divide${x^{18}} + 1$ by (x + 1). We can see the degree of the numerator is 18 and the denominator has a linear equation. So, the degree of quotient starts from 17 and decreases with a difference of one.
$ \Rightarrow \dfrac{{{x^{18}} + 1}}{{x + 1}} = {x^{17}} - {x^{16}} + {x^{15}} - {x^{14}} + .......... + x$
So, the correct answer is “Option d”.
Note: Whenever we face such types of problems we use some important points. Like first of all try to convert series into easy form by using sum of series. While dividing carefully observe the degree of quotient and sign of coefficient. Doing these things will solve all such problems.
Complete step-by-step answer:
As we know, the sum of finite geometric series $1,r,{r^2},{r^3},{r^4}..........{r^{n - 1}}$ with common ratio is given by
$a,ar,a{r^2},a{r^3},a{r^4}..........a{r^{n - 1}} = S = \dfrac{{a({r^n} - 1)}}{{r - 1}}$ = sum (where r is greater than 1)
When r is less than 1 then the formula of sum for n terms of GP will be $\dfrac{{a(1 - {r^n})}}{{1 - r}}$.
Now, $1 + {x^2} + {x^4} + {x^6} + .......... + {x^{34}}$ forms a finite geometric series with common ratio ${x^2}$, first term 1 and number of terms n = 18 (since power of x from 2 to 34 with the difference of 2 is 17 terms and the first term is 1 therefore n = 18) then sum is,
$1 + {x^2} + {x^4} + {x^6} + .......... + {x^{34}} = \dfrac{{{x^{2(18)}} - 1}}{{{x^2} - 1}}$
$1 + {x^2} + {x^4} + {x^6} + .......... + {x^{34}} = \dfrac{{{x^{36}} - 1}}{{{x^2} - 1}}$……….(1)
Now, $1 + x + {x^2} + {x^3} + ........... + {x^{17}}$ forms a finite geometric series with common ratio x, first term 1 and number of terms n = 18 (since power of x from 1 to 17 is 17 terms and first term is 1 therefore n = 18) then sum is
$1 + x + {x^2} + {x^3} + ........... + {x^{17}} = \dfrac{{{x^{18}} - 1}}{{x - 1}}$……………..(2)
Now, Dividing (1) by (2) equation
$ \Rightarrow \dfrac{{\left( {{x^{36}} - 1} \right)\left( {x - 1} \right)}}{{\left( {{x^2} - 1} \right)\left( {{x^{18}} - 1} \right)}} = \dfrac{{{x^{18}} - 1}}{{x + 1}}$
Now, we have to divide${x^{18}} + 1$ by (x + 1). We can see the degree of the numerator is 18 and the denominator has a linear equation. So, the degree of quotient starts from 17 and decreases with a difference of one.
$ \Rightarrow \dfrac{{{x^{18}} + 1}}{{x + 1}} = {x^{17}} - {x^{16}} + {x^{15}} - {x^{14}} + .......... + x$
So, the correct answer is “Option d”.
Note: Whenever we face such types of problems we use some important points. Like first of all try to convert series into easy form by using sum of series. While dividing carefully observe the degree of quotient and sign of coefficient. Doing these things will solve all such problems.
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