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For what value of $n$ is $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$?

 Jun 19, 2019
 #1
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+1

n=1;c=0;a = (n*1i^n);c=c+a;printc,n;n++; if(n<98, goto2, discard=0;

 

 $i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i$?

 

n = 97 =48 + 49i

 

1  =  0 +1 i
2  =  -2 +1 i
3  =  -2 -2 i
4  =  2 -2 i
5  =  2 +3 i
6  =  -4 +3 i
7  =  -4 -4 i
8  =  4 -4 i
9  =  4 +5 i
10  =  -6 +5 i
11  =  -6 -6 i
12  =  6 -6 i
13  =  6 +7 i
14  =  -8 +7 i
15  =  -8 -8 i
16  =  8 -8 i
17  =  8 +9 i
18  =  -10 +9 i
19  =  -10 -10 i
20  =  10 -10 i
21  =  10 +11 i
22  =  -12 +11 i
23  =  -12 -12 i
24  =  12 -12 i
25  =  12 +13 i
26  =  -14 +13 i
27  =  -14 -14 i
28  =  14 -14 i
29  =  14 +15 i
30  =  -16 +15 i
31  =  -16 -16 i
32  =  16 -16 i
33  =  16 +17 i
34  =  -18 +17 i
35  =  -18 -18 i
36  =  18 -18 i
37  =  18 +19 i
38  =  -20 +19 i
39  =  -20 -20 i
40  =  20 -20 i
41  =  20 +21 i
42  =  -22 +21 i
43  =  -22 -22 i
44  =  22 -22 i
45  =  22 +23 i
46  =  -24 +23 i
47  =  -24 -24 i
48  =  24 -24 i
49  =  24 +25 i
50  =  -26 +25 i
51  =  -26 -26 i
52  =  26 -26 i
53  =  26 +27 i
54  =  -28 +27 i
55  =  -28 -28 i
56  =  28 -28 i
57  =  28 +29 i
58  =  -30 +29 i
59  =  -30 -30 i
60  =  30 -30 i
61  =  30 +31 i
62  =  -32 +31 i
63  =  -32 -32 i
64  =  32 -32 i
65  =  32 +33 i
66  =  -34 +33 i
67  =  -34 -34 i
68  =  34 -34 i
69  =  34 +35 i
70  =  -36 +35 i
71  =  -36 -36 i
72  =  36 -36 i
73  =  36 +37 i
74  =  -38 +37 i
75  =  -38 -38 i
76  =  38 -38 i
77  =  38 +39 i
78  =  -40 +39 i
79  =  -40 -40 i
80  =  40 -40 i
81  =  40 +41 i
82  =  -42 +41 i
83  =  -42 -42 i
84  =  42 -42 i
85  =  42 +43 i
86  =  -44 +43 i
87  =  -44 -44 i
88  =  44 -44 i
89  =  44 +45 i
90  =  -46 +45 i
91  =  -46 -46 i
92  =  46 -46 i
93  =  46 +47 i
94  =  -48 +47 i
95  =  -48 -48 i
96  =  48 -48 i
97  =  48 +49 i

 Jun 19, 2019
edited by Guest  Jun 20, 2019
 #2
avatar+102937 
+2

\(i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i\)

 

Note that

 

i + 2i^2 + 3i^3 + 4i^4 =

 

(i + 3i^3) + (4i^4 + 2i^2)  =

 

(i - 3i) + (4 - 2)  =

 

[-2i  + 2]

 

And every  cycle of 4 terms in the summation  will produce   [ -2i + 2 ]

 

So

 

24 cycles of 4 wil produce    24 [ -2i + 2 ]  =   48 - 48i    (1)

 

And the term at the begining of the kth cycle is given by  ( 4k - 3)i^(4k - 3)

 

So....the term at the start of the 25th cycle =  (4 *25 - 3) i ^(4 * 25 - 3) =  97 i^(97)  = 97i      (2)

 

So.....the sum of  (1) and (2)   =   48 - 48i + 97i    =   48 + 49i

 

So 

 

(4*25 - 3)  =  n  =   97

 

Just as the Guest found   !!!!

 

 

cool cool cool

 Jun 20, 2019
edited by CPhill  Jun 20, 2019
edited by CPhill  Jun 20, 2019
edited by CPhill  Jun 20, 2019
 #3
avatar+23041 
+2

For what value of  \(n\)  is  \(i + 2i^2 + 3i^3 + \cdots + ni^n = 48 + 49i\) ?

 

\(\begin{array}{|rcll|} \hline s &=& i + 2i^2 + 3i^3 + \cdots + ni^n \\ \hline \end{array}\)

 

\(\begin{array}{|rclll|} \hline s &=& i + 2i^2 + 3i^3 +4i^4+ \cdots + ni^n \\ is &=& \qquad i^2 +2i^3+3i^4+\ldots (n+1)i^n+ni^{n+1} \\ \hline s-is &=& \underbrace{ i+ i^2+i^3+i^4+\ldots +i^n}_{=S~ (GP)}-ni^{n+1}\\ s(1-i) &=& S-ni^{n+1}\\ &&\begin{array}{|rclll|} \hline S &=& i+ i^2+i^3+i^4+\ldots +i^n \\ iS &=& \qquad i^2+i^3+i^4+\ldots +i^n +i^{n+1} \\ \hline S - iS &=& i-i^{n+1} \\ S (1- i) &=& i( 1-i^{n}) \\ S &=& \dfrac{i( 1-i^{n})}{1-i} \\ \hline \end{array} \\\\ s(1-i) &=& \dfrac{i( 1-i^{n})}{1-i}-ni^{n+1} \\\\ s &=& \dfrac{i( 1-i^{n})}{(1-i)^2}-\dfrac{ni^{n+1}}{ 1-i } \\\\ s &=& \dfrac{i( 1-i^{n})}{(1-i)^2}-\left(\dfrac{ni^{n+1}}{ 1-i }\right)\cdot \left(\dfrac{1-i}{1-i}\right) \\\\ s &=& \dfrac{i( 1-i^{n}) -ni^{n+1}(1-i) }{(1-i)^2} \\\\ s &=& \dfrac{i( 1-i^{n}) -ni^{n}i(1-i) }{(1-i)^2} \\\\ s &=& \dfrac{i\Big( 1-i^{n} -ni^{n}(1-i)\Big) }{(1-i)^2} \quad | \quad (1-i)^2 = -2i \\\\ s &=& \dfrac{i\Big( 1-i^{n} -ni^{n}(1-i)\Big) }{-2i} \\\\ s &=& \dfrac{ 1-i^{n} -ni^{n}(1-i) }{-2 } \\\\ s &=& \dfrac{ -1+i^{n} +ni^{n}(1-i) }{2 } \\\\ s &=& \dfrac{ -1+i^{n} +ni^{n} -ni^{n+1} }{2 } \\\\ \mathbf{s} &=& \mathbf{\dfrac{ i^{n}(1+n) -i^{n+1}n -1 }{2 }} \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \mathbf{s} = \mathbf{\dfrac{ i^{n}(1+n) -i^{n+1}n -1 }{2 }} &=& 48 + 49i \\\\ i^{n}(1+n) -i^{n+1}n -1 &=& 96 + 98i \\\\ \mathbf{{\color{red}i^{n}}(1+n) -{\color{red}i^{n+1}}n } &=& \mathbf{97 + 98i} \\ \hline \end{array} \)

 

We compare the sides of the equation and note that i^n and i^(n+1) are adjacent values.

This means that one is either +i or -i and the other is either +1 or -1.

So there are two solutions for the comparison.

 

1. First possible solution

\(\begin{array}{|rcll|} \hline 98i &=&{ -\color{red}i^{n+1}}n \\ 97 &=& {\color{red}i^{n}}(1+n) \\ \hline \dfrac{98i}{97} &=& \dfrac{-i^{n+1} n}{i^{n} (1+n)} \\\\ \dfrac{98i}{97} &=& \dfrac{-i^{n}i\cdot n}{i^{n} (1+n)} \\\\ \dfrac{98i}{97} &=& \dfrac{- i\cdot n}{ (1+n)} \\\\ \dfrac{98}{97} &=& \dfrac{-n}{ (1+n)} \\\\ (1+n)98 &=& -97n \\ 98+98n &=& -97n \\ 195n &=& -98 \\ 195n &=& -98 \\ n &=& -\dfrac{98}{195} \\ n &=& -0.50256410256 \qquad \text{no solution, n is no integer } \\ \hline \end{array}\)

 

2. Second possible solution

\(\begin{array}{|rcll|} \hline 98i &=& {\color{red}i^{n}}(1+n) \\ 97 &=&{ -\color{red}i^{n+1}}n \\ \hline \dfrac{98i}{97} &=& \dfrac{i^{n} (1+n)}{-i^{n+1} n} \\\\ \dfrac{98i}{97} &=& \dfrac{i^{n} (1+n)}{-i^{n}i\cdot n} \\\\ \dfrac{98i^2}{97} &=& \dfrac{i^{n} (1+n)}{-i^{n} \cdot n} \\\\ \dfrac{-98}{97} &=& \dfrac{1+n}{-n} \\\\ \dfrac{98}{97} &=& \dfrac{1+n}{n} \\\\ 98n &=& 97(1+n) \\ 98n &=& 97 + 97n \\ \mathbf{n} &=& \mathbf{97} \\ \hline \end{array}\)

 

\(i + 2i^2 + 3i^3 + \cdots + 97i^{97} = 48 + 49i\)

 

laugh

 Jun 20, 2019
edited by heureka  Jun 20, 2019

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