Question

The sum of $i-2-3i+4......$upto \[100\] terms, where $i=\sqrt{-1}$ is
$1)50\left( 1-i \right)$
$2)25\left( 1+i \right)$
$3)100\left( 1-i \right)$
$4)25i$

Answer 1

Hint:To solve this question we need to have the knowledge of determining the type of sequence. The next step will be to find the important parameters required to find the sum of the series. In this question we will multiply the whole expression with $i$ and then subtract $Si$ from $S$, hence calculating the sum.

Complete step-by-step solution:
The question asks us to find the sum of the series$i-2-3i+4......$ till \[100\] terms. Now we can see that the sequence given to us is a mixture of arithmetic and geometric progression. In this case we will consider the sum of the equations to be $S$. On writing it mathematically we get:
$\Rightarrow S=i-2-3i+4.......$
The above pattern could be written as:
$\Rightarrow S=i+2{{i}^{2}}+3{{i}^{3}}+4{{i}^{4}}.......100{{i}^{100}}$
The next step to solve this question is to multiply the whole of the terms in LHS and RHS with the $i$, which is a complex number indicating $i=\sqrt{-1}$. So on multiplying we get the term $Si$ equal to:
$\Rightarrow Si=\left( i+2{{i}^{2}}+3{{i}^{3}}+4{{i}^{4}}.......100{{i}^{100}} \right)i$
On multiplying $i$ with each term we get:
$\Rightarrow Si={{i}^{2}}+2{{i}^{3}}+3{{i}^{4}}+4{{i}^{5}}.......100{{i}^{101}}$
The power of $i$in each term increases by $1$, because of the same base, which is $i$.
Now we got the 2 equations as $Si$ and $S$. We will be subtracting $Si$ from $S$ which will help us in getting the following result:
$\Rightarrow S-Si=i+2{{i}^{2}}+3{{i}^{3}}+4{{i}^{4}}.......100{{i}^{100}}-\left( {{i}^{2}}+2{{i}^{3}}+3{{i}^{4}}+4{{i}^{5}}.......100{{i}^{101}} \right)$
$\Rightarrow S-Si=i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}.......{{i}^{100}}-100{{i}^{101}}$
The above series is in geometric progression. As each term is increasing as the multiple of $i$.
So we will be applying a formula for geometric progression in that Series to find the sum of the series.
$\Rightarrow S-Si=[i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}.......{{i}^{100}}]-100{{i}^{101}}$
On calculating the sum we get:
$\Rightarrow S\left( 1-i \right)=\dfrac{i\left( {{i}^{100}}-1 \right)}{i-1}-100{{i}^{101}}$
$\Rightarrow S\left( 1-i \right)=-100{{i}^{101}}$
$\Rightarrow S=\dfrac{-100{{i}^{101}}}{1-i}$
We will now rationalise the term so that the complex term from the denominator is removed. On doing this we get:
$\Rightarrow S=\dfrac{-100{{i}^{101}}\left( 1+i \right)}{\left( 1-i \right)\left( 1+i \right)}$
$\Rightarrow S=\dfrac{-100{{i}^{101}}\left( 1+i \right)}{\left( 1-{{i}^{2}} \right)}$
$\Rightarrow S=\dfrac{-100{{i}^{101}}\left( 1+i \right)}{\left( 1-(-1) \right)}$
$\Rightarrow S=\dfrac{-100{{i}^{101}}\left( 1+i \right)}{2}$
The power of $i$ is $101$ which can be written as a sum of $100$ and $1$ . Since ${{i}^{100}}=1$, so the expression becomes:
$\Rightarrow S=-50i\left( 1+i \right)$
$\Rightarrow S=-50\left( i+{{i}^{2}} \right)$
$\Rightarrow S=-50\left( i-1 \right)$
$\Rightarrow S=50\left( 1-i \right)$
$\therefore $ The sum of $i-2-3i+4......$upto \[100\] terms, where $i=\sqrt{-1}$ is $1)50\left( 1-i \right)$

Note:To solve questions like this the first step should be to find the type of series given in the question. To find this we will be finding either the common difference or common ratio so that we know about the sequence.