What is the value of the sum \(\mathop \sum \limits_{{\rm{n}} = 2}^{11} \left( {{{\rm{i}}^{\rm{n}}} + {{\rm{i}}^{{\rm{n}} + 1}}} \right),\) where \({\rm{i}} = \sqrt { - 1} ?\)
-2i
Evaluating the Sum of Complex Number Series
The question asks for the value of the sum of a series involving powers of the imaginary unit, \({\rm{i}}\), defined as \({\rm{i}} = \sqrt { - 1}\). The sum is given by:
\(\mathop \sum \limits_{{\rm{n}} = 2}^{11} \left( {{{\rm{i}}^{\rm{n}}} + {{\rm{i}}^{{\rm{n}} + 1}}} \right)\)
We can separate the sum into two parts:
\(\mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{\rm{n}}} + \mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{{\rm{n}} + 1}}\)
Let's evaluate each part separately.
Understanding Powers of the Imaginary Unit i
The powers of \({\rm{i}}\) follow a repeating pattern:
- \({\rm{i}}^1 = {\rm{i}}\)
- \({\rm{i}}^2 = -1\)
- \({\rm{i}}^3 = {\rm{i}}^2 \cdot {\rm{i}} = -1 \cdot {\rm{i}} = -{\rm{i}}\)
- \({\rm{i}}^4 = {\rm{i}}^2 \cdot {\rm{i}}^2 = (-1) \cdot (-1) = 1\)
- \({\rm{i}}^5 = {\rm{i}}^4 \cdot {\rm{i}} = 1 \cdot {\rm{i}} = {\rm{i}}\) (The pattern repeats)
A key property is that the sum of four consecutive powers of \({\rm{i}}\) is always zero:
\({\rm{i}}^{\rm{k}} + {\rm{i}}^{{\rm{k}} + 1} + {\rm{i}}^{{\rm{k}} + 2} + {\rm{i}}^{{\rm{k}} + 3} = {\rm{i}}^{\rm{k}}(1 + {\rm{i}} + {\rm{i}}^2 + {\rm{i}}^3) = {\rm{i}}^{\rm{k}}(1 + {\rm{i}} - 1 - {\rm{i}}) = {\rm{i}}^{\rm{k}}(0) = 0\)
Evaluating the First Sum: \(\mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{\rm{n}}}\)
This sum includes terms from \({\rm{n}} = 2\) to \({\rm{n}} = 11\). The number of terms is \(11 - 2 + 1 = 10\). The terms are:
\({{\rm{i}}^{\rm{2}}} + {{\rm{i}}^{\rm{3}}} + {{\rm{i}}^{\rm{4}}} + {{\rm{i}}^{\rm{5}}} + {{\rm{i}}^{\rm{6}}} + {{\rm{i}}^{\rm{7}}} + {{\rm{i}}^{\rm{8}}} + {{\rm{i}}^{\rm{9}}} + {{\rm{i}}^{{\rm{10}}}} + {{\rm{i}}^{{\rm{11}}}}\)
Let's write out their values based on the cycle:
\(-1 + (-{\rm{i}}) + 1 + {\rm{i}} + (-1) + (-{\rm{i}}) + 1 + {\rm{i}} + (-1) + (-{\rm{i}})\)
We can group these 10 terms into sets of 4, followed by the remaining terms:
\((-1 - {\rm{i}} + 1 + {\rm{i}}) + (-1 - {\rm{i}} + 1 + {\rm{i}}) + (-1 - {\rm{i}})\)
Using the property that the sum of 4 consecutive powers is 0:
\(0 + 0 + (-1 - {\rm{i}}) = -1 - {\rm{i}}\)
So, \(\mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{\rm{n}}} = -1 - {\rm{i}}\).
Evaluating the Second Sum: \(\mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{{\rm{n}} + 1}}\)
This sum includes terms from \({\rm{n}} = 2\) to \({\rm{n}} = 11\), which corresponds to powers from \({\rm{n}} + 1 = 3\) to \({\rm{n}} + 1 = 12\). The number of terms is \(11 - 2 + 1 = 10\). The terms are:
\({{\rm{i}}^{\rm{3}}} + {{\rm{i}}^{\rm{4}}} + {{\rm{i}}^{\rm{5}}} + {{\rm{i}}^{\rm{6}}} + {{\rm{i}}^{\rm{7}}} + {{\rm{i}}^{\rm{8}}} + {{\rm{i}}^{\rm{9}}} + {{\rm{i}}^{{\rm{10}}}} + {{\rm{i}}^{{\rm{11}}}} + {{\rm{i}}^{{\rm{12}}}}\)
Let's write out their values:
\(-{\rm{i}} + 1 + {\rm{i}} + (-1) + (-{\rm{i}}) + 1 + {\rm{i}} + (-1) + (-{\rm{i}}) + 1\)
We can group these 10 terms into sets of 4, followed by the remaining terms:
\(( - {\rm{i}} + 1 + {\rm{i}} - 1) + (-{\rm{i}} + 1 + {\rm{i}} - 1) + (-{\rm{i}} + 1)\)
Using the property that the sum of 4 consecutive powers is 0:
\(0 + 0 + (-{\rm{i}} + 1) = 1 - {\rm{i}}\)
So, \(\mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{{\rm{n}} + 1}} = 1 - {\rm{i}}\).
Combining the Results
The total sum is the sum of the two parts we evaluated:
\(\left( \mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{\rm{n}}} \right) + \left( \mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{{\rm{n}} + 1}} \right) = (-1 - {\rm{i}}) + (1 - {\rm{i}})\)
\(= -1 - {\rm{i}} + 1 - {\rm{i}}\)
Combining the real and imaginary parts:
\(= (-1 + 1) + (-{\rm{i}} - {\rm{i}})\)
\(= 0 - 2{\rm{i}}\)
\(= -2{\rm{i}}\)
Alternative Approach: Factoring
We can also factor the term inside the summation:
\({{\rm{i}}^{\rm{n}}} + {{\rm{i}}^{{\rm{n}} + 1}} = {{\rm{i}}^{\rm{n}}}(1 + {\rm{i}})\)
The sum becomes:
\(\mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{\rm{n}}}(1 + {\rm{i}})\)
Since \((1 + {\rm{i}})\) is constant with respect to \({\rm{n}}\), we can pull it out of the summation:
\((1 + {\rm{i}}) \mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{\rm{n}}}\)
We already calculated \(\mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{\rm{n}}} = -1 - {\rm{i}}\).
So the sum is:
\((1 + {\rm{i}})(-1 - {\rm{i}})\)
We can factor out -1 from the second term:
\((1 + {\rm{i}})(-1(1 + {\rm{i}})) = -(1 + {\rm{i}})^2\)
Now, expand \((1 + {\rm{i}})^2\):
\((1 + {\rm{i}})^2 = 1^2 + 2(1)({\rm{i}}) + {\rm{i}}^2 = 1 + 2{\rm{i}} + (-1) = 1 + 2{\rm{i}} - 1 = 2{\rm{i}}\)
Substitute this back into the expression:
\(-(1 + {\rm{i}})^2 = -(2{\rm{i}}) = -2{\rm{i}}\)
Both approaches yield the same result.
| Term | Value of \({\rm{i}}^{\rm{n}}\) | Value of \({\rm{i}}^{{\rm{n}}+1}\) | Value of \({\rm{i}}^{\rm{n}} + {\rm{i}}^{{\rm{n}}+1}\) |
|---|---|---|---|
| n=2 | \({\rm{i}}^2 = -1\) | \({\rm{i}}^3 = -{\rm{i}}\) | \(-1 - {\rm{i}}\) |
| n=3 | \({\rm{i}}^3 = -{\rm{i}}\) | \({\rm{i}}^4 = 1\) | \(-{\rm{i}} + 1\) |
| n=4 | \({\rm{i}}^4 = 1\) | \({\rm{i}}^5 = {\rm{i}}\) | \(1 + {\rm{i}}\) |
| n=5 | \({\rm{i}}^5 = {\rm{i}}\) | \({\rm{i}}^6 = -1\) | \({\rm{i}} - 1\) |
| n=6 | \({\rm{i}}^6 = -1\) | \({\rm{i}}^7 = -{\rm{i}}\) | \(-1 - {\rm{i}}\) |
| n=7 | \({\rm{i}}^7 = -{\rm{i}}\) | \({\rm{i}}^8 = 1\) | \(-{\rm{i}} + 1\) |
| n=8 | \({\rm{i}}^8 = 1\) | \({\rm{i}}^9 = {\rm{i}}\) | \(1 + {\rm{i}}\) |
| n=9 | \({\rm{i}}^9 = {\rm{i}}\) | \({\rm{i}}^{10} = -1\) | \({\rm{i}} - 1\) |
| n=10 | \({\rm{i}}^{10} = -1\) | \({\rm{i}}^{11} = -{\rm{i}}\) | \(-1 - {\rm{i}}\) |
| n=11 | \({\rm{i}}^{11} = -{\rm{i}}\) | \({\rm{i}}^{12} = 1\) | \(-{\rm{i}} + 1\) |
Summing the last column:
\((-1 - {\rm{i}}) + (-{\rm{i}} + 1) + (1 + {\rm{i}}) + ({\rm{i}} - 1) + (-1 - {\rm{i}}) + (-{\rm{i}} + 1) + (1 + {\rm{i}}) + ({\rm{i}} - 1) + (-1 - {\rm{i}}) + (-{\rm{i}} + 1)\)
Group terms:
\((-1+1+1-1-1+1+1-1-1+1) + (-{\rm{i}}-{\rm{i}}+{\rm{i}}+{\rm{i}}-{\rm{i}}-{\rm{i}}+{\rm{i}}+{\rm{i}}-{\rm{i}}-{\rm{i}})\)
\((0) + (-4{\rm{i}})\)
\(= -4{\rm{i}}\)
Wait, let's re-examine the terms of the sum \(i^n + i^{n+1} = i^n(1+i)\). The values of \(1+i\) are constant. We are summing \(i^n\) for n=2 to 11, and multiplying the result by \(1+i\).
Sum of \(i^n\) for n=2 to 11 is \(i^2 + i^3 + i^4 + i^5 + i^6 + i^7 + i^8 + i^9 + i^{10} + i^{11}\)
\(= (-1) + (-i) + (1) + (i) + (-1) + (-i) + (1) + (i) + (-1) + (-i)\)
\(= (-1 - i + 1 + i) + (-1 - i + 1 + i) + (-1 - i)\)
\(= 0 + 0 + (-1 - i)\)
\(= -1 - i\)
The total sum is \((1+i) \times (-1-i)\)
\(= -(1+i)(1+i) = -(1^2 + 2(1)(i) + i^2) = -(1 + 2i - 1) = -(2i) = -2i\)
The calculation is correct. The table sum was miscalculated manually. Let's sum the last column again.
Sum \(= (-1 - {\rm{i}}) + (1 - {\rm{i}}) + (1 + {\rm{i}}) + (-1 + {\rm{i}}) + (-1 - {\rm{i}}) + (1 - {\rm{i}}) + (1 + {\rm{i}}) + (-1 + {\rm{i}}) + (-1 - {\rm{i}}) + (1 - {\rm{i}})\)
Group in pairs of 4 terms (which correspond to a cycle of 4 values for \(i^n\)):
\((-1 - {\rm{i}}) + (1 - {\rm{i}}) + (1 + {\rm{i}}) + (-1 + {\rm{i}})\) for n=2,3,4,5 sum is \((-1+1+1-1) + (-i-i+i+i) = 0 + 0 = 0\).
The next 4 terms (n=6,7,8,9): \((-1 - {\rm{i}}) + (1 - {\rm{i}}) + (1 + {\rm{i}}) + (-1 + {\rm{i}})\) sum is also 0.
The last 2 terms (n=10,11): \((-1 - {\rm{i}}) + (1 - {\rm{i}})\) sum is \((-1+1) + (-i-i) = 0 - 2i = -2i\).
Total sum = \(0 + 0 + (-2{\rm{i}}) = -2{\rm{i}}\).
This confirms the result obtained by factoring.
Final Answer Summary
The value of the sum \(\mathop \sum \limits_{{\rm{n}} = 2}^{11} \left( {{{\rm{i}}^{\rm{n}}} + {{\rm{i}}^{{\rm{n}} + 1}}} \right)\) is \(-2{\rm{i}}\).
Revision Table: Sum of Complex Numbers Series
| Concept | Description | Key Property |
|---|---|---|
| Complex Number \({\rm{i}}\) | The imaginary unit, \({\rm{i}} = \sqrt{-1}\). | \({\rm{i}}^2 = -1\) |
| Powers of \({\rm{i}}\) | \({\rm{i}}^1={\rm{i}}, {\rm{i}}^2=-1, {\rm{i}}^3=-{\rm{i}}, {\rm{i}}^4=1\). The pattern repeats every 4 powers. | \({\rm{i}}^{\rm{k}} = {\rm{i}}^{{\rm{k}} \pmod 4}\) (using 1,2,3,0 or 1,2,3,4 in the exponent) |
| Sum of Consecutive Powers of \({\rm{i}}\) | The sum of any four consecutive powers of \({\rm{i}}\) is 0. | \({\rm{i}}^{\rm{k}} + {\rm{i}}^{{\rm{k}}+1} + {\rm{i}}^{{\rm{k}}+2} + {\rm{i}}^{{\rm{k}}+3} = 0\) |
| Summation Notation | \(\mathop \sum \limits_{{\rm{n}} = {\rm{a}}}^{{\rm{b}}} {\rm{f}}({\rm{n}})\) means summing the expression \({\rm{f}}({\rm{n}})\) for integer values of \({\rm{n}}\) from \({\rm{a}}\) to \({\rm{b}}\). | Number of terms = \({\rm{b}} - {\rm{a}} + 1\) |
Additional Information on Complex Numbers and Series
Complex numbers are numbers of the form \({\rm{a}} + {\rm{b}}{\rm{i}}\), where \({\rm{a}}\) and \({\rm{b}}\) are real numbers, and \({\rm{i}}\) is the imaginary unit. They are fundamental in many areas of mathematics, physics, and engineering.
Series involving complex numbers, like the one in this problem, are common in topics such as Fourier series and power series in complex analysis. Understanding the cyclic nature of powers of \({\rm{i}}\) is crucial for evaluating such sums efficiently.
A finite geometric series has the form \(\mathop \sum \limits_{{\rm{k}} = 0}^{{\rm{N}} - 1} {\rm{ar}}^{\rm{k}}\). Its sum is given by \(\frac{{\rm{a}}(1 - {\rm{r}}^{\rm{N}})}{1 - {\rm{r}}}\), provided \({\rm{r}} \ne 1\).
In our problem, \(\mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{\rm{n}}}\) is a geometric series with \({\rm{a}} = {\rm{i}}^2 = -1\), \({\rm{r}} = {\rm{i}}\), and the number of terms is \({\rm{N}} = 10\). The sum would be \(\frac{(-1)(1 - {\rm{i}}^{10})}{1 - {\rm{i}}}\). \({\rm{i}}^{10} = {\rm{i}}^{8} \cdot {\rm{i}}^2 = (1) \cdot (-1) = -1\). So the sum is \(\frac{(-1)(1 - (-1))}{1 - {\rm{i}}} = \frac{(-1)(2)}{1 - {\rm{i}}} = \frac{-2}{1 - {\rm{i}}}\). To simplify, multiply numerator and denominator by the conjugate of the denominator, which is \(1 + {\rm{i}}\): \(\frac{-2}{1 - {\rm{i}}} \times \frac{1 + {\rm{i}}}{1 + {\rm{i}}} = \frac{-2(1 + {\rm{i}})}{1^2 - {\rm{i}}^2} = \frac{-2 - 2{\rm{i}}}{1 - (-1)} = \frac{-2 - 2{\rm{i}}}{2} = -1 - {\rm{i}}\). This confirms our previous calculation of \(\mathop \sum \limits_{{\rm{n}} = 2}^{11} {{\rm{i}}^{\rm{n}}}\) using the geometric series formula as well.
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