Question

$\text {solve \displaystyle \sum \limits^{100}_{r = 1} \: \: i {}^{2r + 1} }$

I found the answer but not confirm about the method. My answer was 0 please solve it using best method.

This is complex numbers​

Answer 1

$\texttt{\textsf{\large{\underline{Solution}:}}}$

Given To Evaluate:

$= \displaystyle \rm \sum_{r = 1}^{100} {i}^{2r + 1}$

The sum will be equal to:

$\rm = {i}^{3} + {i}^{5} + ... + {i}^{201}$

Now consider the first two terms:

$\rm = {i}^{3} + {i}^{5}$

$\rm = {i}^{4 \times 0 + 3} + {i}^{4 \times 1 + 1}$

$\rm = - i+i$

$\rm =0$

Again, consider the next two terms:

$\rm = {i}^{7} + {i}^{9}$

$\rm = {i}^{4 \times 1 + 3} + {i}^{4 \times 2 + 1}$

$\rm = - i +i$

$\rm =0$

Now, this pattern will continue till the last term i.e.,

$\rm = {i}^{199} + {i}^{201}$

$\rm =0$

Therefore:

$: \longmapsto \displaystyle \rm \sum_{r = 1}^{100} {i}^{2r + 1} = 0$

Which is our required answer.

$\texttt{\textsf{\large{\underline{Additional Information}:}}}$

$\rm1.\ i^{4n} = 1$

$\rm2. \ i^{4n+1} = i$

$\rm3.\ i^{4n+2} = -1$

$\rm4.\ i^{4n+3} = -i$

Answer 2

To Solve :-

$\displaystyle \sum \limits^{100}_{r = 1} \: \: i {}^{2r + 1}$

Solution :-

In the given question, the concept of complex numbers is use. iota is a complex number which is equal to √-1 and i² = -1. iota raise to the power multiple of 4 is 1. We will use this concept to solve the given problem.

The expansion of given expression is given by,

$\longrightarrow \displaystyle \sum \limits^{100}_{r = 1} \: \: i {}^{2r + 1} \: = ( {i}^{ {2(1) + 1}}) + ( {i}^{2(2) + 1} ) + ( {i}^{2(3) + 1} ) +( {i}^{2(4) + 1}) + ... + ( {i}^{2(99 ) + 1} ) + ( {i}^{2(100) + 1})$

$\longrightarrow \displaystyle \sum \limits^{100}_{r = 1} \: \: i {}^{2r + 1} \: = ( {i}^{ {2 + 1}}) + ( {i}^{{4 + 1}} ) + ( {i}^{6 + 1} ) +( {i}^{8 + 1})+ ... + ( {i}^{198 + 1} ) + ( {i}^{200 + 1})$

$\longrightarrow \displaystyle \sum \limits^{100}_{r = 1} \: \: i {}^{2r + 1} \: = ( {i}^{ {3}}) + ( {i}^{{5}} ) + ( {i}^{7} ) +( {i}^{9}) + ... + ( {i}^{199} ) + ( {i}^{201})$

$\longrightarrow \displaystyle \sum \limits^{100}_{r = 1} \: \: i {}^{2r + 1} \: = ( {i}^{ {2}}.i) + ( {i}^{{4}}.i ) + ( {i}^{4}. {i}^{2}.i ) +( {i}^{8}.i) +... + ( {i}^{196}.i^{2}.i ) + ( {i}^{200}.i)$

Now substitute,

• iota raise to the power multiple of 4 = 1
• iota raise to the power 2 = -1

$\longrightarrow \displaystyle \sum \limits^{100}_{r = 1} \: \: i {}^{2r + 1} \: = ( - i) + (i ) + ( - i ) +( i)+ ... + ( - i ) + (i)$

Now, we can examine a pattern involved which each term of the series , as -i + i - i + i +. . .- i + i.

All the iota terms will be cancelled out and the final answer will be 0.

Hence,

$\displaystyle \sum \limits^{100}_{r = 1} \: \: i {}^{2r + 1} = 0$

Your answer was correct :)