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Difference between revisions of "1964 AHSME Problems/Problem 34"

(Created page with "==Problem== If <math>n</math> is a multiple of <math>4</math>, the sum <math>s=1+2i+3i^2+\cdots+(n+1)i^n</math>, where <math>i=\sqrt{-1}</math>, equals: <math>\textbf{(A) }1+i\...")
 
(→‎Solution)
 
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If <math>n</math> is a multiple of <math>4</math>, the sum <math>s=1+2i+3i^2+\cdots+(n+1)i^n</math>, where <math>i=\sqrt{-1}</math>, equals:
 
If <math>n</math> is a multiple of <math>4</math>, the sum <math>s=1+2i+3i^2+\cdots+(n+1)i^n</math>, where <math>i=\sqrt{-1}</math>, equals:
  
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<math>\textbf{(A) }1+i\qquad\textbf{(B) }\frac{1}{2}(n+2)\qquad\textbf{(C) }\frac{1}{2}(n+2-ni)\qquad\textbf{(D) }\frac{1}{2}[(n+1)(1-i)+2]\qquad \textbf{(E) }\frac{1}{8}(n^2+8-4ni)</math>
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<math>\textbf{(A) }1+i\qquad\textbf{(B) }\frac{1}{2}(n+2)\qquad\textbf{(C) }\frac{1}{2}(n+2-ni)\qquad</math>
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<math>\textbf{(D) }\frac{1}{2}[(n+1)(1-i)+2]\qquad \textbf{(E) }\frac{1}{8}(n^2+8-4ni)</math>
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==Solution==
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The real part is <math>1 + 3i^2 + 5i^4 + ...</math>, which is <math>1 - 3 + 5 - 7 + ...</math>.  If <math>n</math> is a multiple of <math>4</math>, then we have an odd number of terms in total:  we start with <math>1</math> at <math>n=0</math>, then add two more terms to get <math>1 - 3 + 5</math> at <math>n=4</math>, etc.  With each successive addition, we're really adding a total of <math>2</math>, since <math>-3 + 5 = 2</math>, and <math>-7 + 9 = 2</math>, etc. 
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At <math>n = 0</math>, the sum is <math>1</math>, and at <math>n=4</math>, the sum is <math>3</math>.  Since the sum increases linearly, the real part of the sum is <math>1 + \frac{1}{2}n</math>.
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The imaginary part is <math>\frac{1}{i}(2i + 4i^3 + 6i^5 + ...)</math>, which is <math>2 - 4 + 6 - 8 + ...</math>.  This time, we have an even number of terms.  We group pairs of terms to get <math>(2-4) + (6-8) + ...</math>, and notice that each pair gives <math>-2</math>.  Again, with <math>n=0</math> the imaginary part is <math>0</math>, while with <math>n=4</math> the imaginary part is <math>-2</math>.  Since again the sum increases linearly, this means the imaginary part is <math>-\frac{n}{2}</math>.
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Combining the real and imaginary parts gives <math>\frac{n}{2} + 1 - \frac{n}{2}i</math>, which is equivalent to option <math>\boxed{\textbf{(C)}}</math>.
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==See Also==
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{{AHSME 40p box|year=1964|num-b=33|num-a=35}}
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[[Category:Introductory Algebra Problems]]
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{{MAA Notice}}

Latest revision as of 01:49, 25 July 2019

Problem

If $n$ is a multiple of $4$, the sum $s=1+2i+3i^2+\cdots+(n+1)i^n$, where $i=\sqrt{-1}$, equals:

$\textbf{(A) }1+i\qquad\textbf{(B) }\frac{1}{2}(n+2)\qquad\textbf{(C) }\frac{1}{2}(n+2-ni)\qquad$

$\textbf{(D) }\frac{1}{2}[(n+1)(1-i)+2]\qquad \textbf{(E) }\frac{1}{8}(n^2+8-4ni)$

Solution

The real part is $1 + 3i^2 + 5i^4 + ...$, which is $1 - 3 + 5 - 7 + ...$. If $n$ is a multiple of $4$, then we have an odd number of terms in total: we start with $1$ at $n=0$, then add two more terms to get $1 - 3 + 5$ at $n=4$, etc. With each successive addition, we're really adding a total of $2$, since $-3 + 5 = 2$, and $-7 + 9 = 2$, etc.

At $n = 0$, the sum is $1$, and at $n=4$, the sum is $3$. Since the sum increases linearly, the real part of the sum is $1 + \frac{1}{2}n$.

The imaginary part is $\frac{1}{i}(2i + 4i^3 + 6i^5 + ...)$, which is $2 - 4 + 6 - 8 + ...$. This time, we have an even number of terms. We group pairs of terms to get $(2-4) + (6-8) + ...$, and notice that each pair gives $-2$. Again, with $n=0$ the imaginary part is $0$, while with $n=4$ the imaginary part is $-2$. Since again the sum increases linearly, this means the imaginary part is $-\frac{n}{2}$.

Combining the real and imaginary parts gives $\frac{n}{2} + 1 - \frac{n}{2}i$, which is equivalent to option $\boxed{\textbf{(C)}}$.

See Also

1964 AHSC (Problems • Answer Key • Resources)
Preceded by
Problem 33 		Followed by
Problem 35
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 • 31 • 32 • 33 • 34 • 35 • 36 • 37 • 38 • 39 • 40
All AHSME Problems and Solutions

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