Question

If $\sum\limits_{k = 0}^{100} {{i^k}} = x + iy$, then the values of $x$ and $y$ are
A. $x = - 1,y = 0$
B. $x = 1,y = 1$
C. $x = 1,y = 0$
D. $x = 0,y = 1$

Answer 1

Hint: First, we shall analyze the given information so that we are able to solve the given problem. Here, we are given that $\sum\limits_{k = 0}^{100} {{i^k}} = x + iy$
We are asked to calculate the values of $x$ and $y$. We need to substitute the values of $k$ from $0$ to $100$ in the given equation
We will note that the sequence is in geometric progression. Then we need to apply the formula of the sum of ${n^{th}}$ terms of G.P
Formula to be used:
The formula to calculate the sum of ${n^{th}}$terms of G.P is as follows.
${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},r > 1$ where$a$ is the first term of G.P,$r$is the common ratio of G.P, and is$n$ the number of terms.

Complete step by step answer:
It is given that$\sum\limits_{k = 0}^{100} {{i^k}} = x + iy$
Now, we shall substitute the values of $k$ from $0$ to $100$ in the given equation.
That is ${i^0} + {i^1} + {i^{2 + }}........ + {i^{100}} = x + iy$
$ \Rightarrow 1 + {i^1} + {i^{2 + }}........ + {i^{100}} = x + iy$……$\left( 1 \right)$
We can note that the above series is in geometric series, where the first term$a = 1$, the common ratio $r = i$
Number of terms $n = 101$
Now, we shall apply the formula of the sum of ${n^{th}}$ terms of G.P
${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}}$ , where $r \ne 1$
Hence, the equation $\left( 1 \right)$ becomes
$\dfrac{{1\left( {1 - {i^{101}}} \right)}}{{1 - i}} = x + iy$ ……$\left( 2 \right)$
Now,${i^{101}} = \left( {{i^{100}}} \right) \times i$
$ = {\left( {{i^4}} \right)^{25}} \times i$
$ = i$
Hence $\left( 2 \right)$ becomes,
$1\dfrac{{\left( {1 - i} \right)}}{{1 - i}} = x + iy$
$ \Rightarrow \dfrac{{1 - i}}{{1 - i}} = x + iy$
$ \Rightarrow x + iy = 1$
$ \Rightarrow x + iy = 1 + oi$
Now we shall compare the real part and the imaginary part on both sides.
Hence, we get$x = 1,y = 0$

So, the correct answer is “Option C”.

Note: If we are given the sum of ${n^{th}}$ terms of G.P, then we need to apply the formula${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},r > 1$ and${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}},r \ne 1$ where$a$ is the first term of G.P,$r$is the common ratio of G.P and is$n$ the number of terms.
Since the common ratio is greater than one, we applied the formula${S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},r > 1$
Suppose the common ratio is smaller than one, we need to apply the formula${S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}},r \ne 1$