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Question: If \(\sum\limits_{k = 0}^{100} {{i^k}} = x + iy\), then the values of \(x\) and \(y\) are A. \(x ...

If ∑k=0100ik=x+iy\sum\limits_{k = 0}^{100} {{i^k}} = x + iy, then the values of xx and yy are
A. x=−1,y=0x = - 1,y = 0
B. x=1,y=1x = 1,y = 1
C. x=1,y=0x = 1,y = 0
D. x=0,y=1x = 0,y = 1

Explanation

Solution

First, we shall analyze the given information so that we are able to solve the given problem. Here, we are given that ∑k=0100ik=x+iy\sum\limits_{k = 0}^{100} {{i^k}} = x + iy
We are asked to calculate the values of xx and yy. We need to substitute the values of kk from 00 to 100100 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 nth{n^{th}} terms of G.P
Formula to be used:
The formula to calculate the sum of nth{n^{th}}terms of G.P is as follows.
Sn=a(rn−1)r−1,r>1{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},r > 1 whereaa is the first term of G.P,rris the common ratio of G.P, and isnn the number of terms.

Complete step by step answer:
It is given that∑k=0100ik=x+iy\sum\limits_{k = 0}^{100} {{i^k}} = x + iy
Now, we shall substitute the values of kk from 00 to 100100 in the given equation.
That is i0+i1+i2+........+i100=x+iy{i^0} + {i^1} + {i^{2 + }}........ + {i^{100}} = x + iy
⇒1+i1+i2+........+i100=x+iy\Rightarrow 1 + {i^1} + {i^{2 + }}........ + {i^{100}} = x + iy……(1)\left( 1 \right)
We can note that the above series is in geometric series, where the first terma=1a = 1, the common ratio r=ir = i
Number of terms n=101n = 101
Now, we shall apply the formula of the sum of nth{n^{th}} terms of G.P
Sn=a(1−rn)1−r{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}} , where r≠1r \ne 1
Hence, the equation (1)\left( 1 \right) becomes
1(1−i101)1−i=x+iy\dfrac{{1\left( {1 - {i^{101}}} \right)}}{{1 - i}} = x + iy ……(2)\left( 2 \right)
Now,i101=(i100)×i{i^{101}} = \left( {{i^{100}}} \right) \times i
=(i4)25×i= {\left( {{i^4}} \right)^{25}} \times i
=i= i
Hence (2)\left( 2 \right) becomes,
1(1−i)1−i=x+iy1\dfrac{{\left( {1 - i} \right)}}{{1 - i}} = x + iy
⇒1−i1−i=x+iy\Rightarrow \dfrac{{1 - i}}{{1 - i}} = x + iy
⇒x+iy=1\Rightarrow x + iy = 1
⇒x+iy=1+oi\Rightarrow x + iy = 1 + oi
Now we shall compare the real part and the imaginary part on both sides.
Hence, we getx=1,y=0x = 1,y = 0

So, the correct answer is “Option C”.

Note: If we are given the sum of nth{n^{th}} terms of G.P, then we need to apply the formulaSn=a(rn−1)r−1,r>1{S_n} = \dfrac{{a\left( {{r^n} - 1} \right)}}{{r - 1}},r > 1 andSn=a(1−rn)1−r,r≠1{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}},r \ne 1 whereaa is the first term of G.P,rris the common ratio of G.P and isnn the number of terms.
Since the common ratio is greater than one, we applied the formulaSn=a(rn−1)r−1,r>1{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 formulaSn=a(1−rn)1−r,r≠1{S_n} = \dfrac{{a\left( {1 - {r^n}} \right)}}{{1 - r}},r \ne 1