Question:
The value of $\sum\limits^{n}_{k=0}\left(i^{k}+i^{k+1}\right)$, where $i^2 = -1$, is equal to
The value of $\sum\limits^{n}_{k=0}\left(i^{k}+i^{k+1}\right)$, where $i^2 = -1$, is equal to
Updated On: Apr 8, 2024
- $i-i^n$
- $-i-i^{n+1}$
- $i-i^{n+1}$
- $i-i^{n+2}$
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The Correct Option is D
Solution and Explanation
$\displaystyle\sum_{k=0}^{n}\left(i^{k}+i^{k+1}\right)=\displaystyle\sum_{k=0}^{n} i^{k}(i+1)$
$=(i+1)\left[1+i^{1}+\ldots+i^{n}\right]$
$=(i+1)\left[\frac{1\left(1-i^{n+1}\right)}{(1-i)}\right] \times \frac{1+i}{1+i}$
$=\frac{(1+i)^{2}\left[1-i^{n+1}\right]}{1^{2}+1}=\frac{(1-1+2 i)\left(1-i^{n+1}\right)}{2}$
$=\frac{2 i}{2}\left(1-i^{n+1}\right)=i-i^{n+2}$
$=(i+1)\left[1+i^{1}+\ldots+i^{n}\right]$
$=(i+1)\left[\frac{1\left(1-i^{n+1}\right)}{(1-i)}\right] \times \frac{1+i}{1+i}$
$=\frac{(1+i)^{2}\left[1-i^{n+1}\right]}{1^{2}+1}=\frac{(1-1+2 i)\left(1-i^{n+1}\right)}{2}$
$=\frac{2 i}{2}\left(1-i^{n+1}\right)=i-i^{n+2}$
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Concepts Used:
Complex Numbers and Quadratic Equations
Complex Number: Any number that is formed as a+ib is called a complex number. For example: 9+3i,7+8i are complex numbers. Here i = -1. With this we can say that i² = 1. So, for every equation which does not have a real solution we can use i = -1.
Quadratic equation: A polynomial that has two roots or is of the degree 2 is called a quadratic equation. The general form of a quadratic equation is y=ax²+bx+c. Here a≠0, b and c are the real numbers.
