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

The sum of i−2−3i+4......i-2-3i+4......upto 100100 terms, where i=−1i=\sqrt{-1} is
1)50(1−i)1)50\left( 1-i \right)
2)25(1+i)2)25\left( 1+i \right)
3)100(1−i)3)100\left( 1-i \right)
4)25i4)25i

Explanation

Solution

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 ii and then subtract SiSi from SS, hence calculating the sum.

Complete step-by-step solution:
The question asks us to find the sum of the seriesi−2−3i+4......i-2-3i+4...... till 100100 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 SS. On writing it mathematically we get:
⇒S=i−2−3i+4.......\Rightarrow S=i-2-3i+4.......
The above pattern could be written as:
⇒S=i+2i2+3i3+4i4.......100i100\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 ii, which is a complex number indicating i=−1i=\sqrt{-1}. So on multiplying we get the term SiSi equal to:
⇒Si=(i+2i2+3i3+4i4.......100i100)i\Rightarrow Si=\left( i+2{{i}^{2}}+3{{i}^{3}}+4{{i}^{4}}.......100{{i}^{100}} \right)i
On multiplying ii with each term we get:
⇒Si=i2+2i3+3i4+4i5.......100i101\Rightarrow Si={{i}^{2}}+2{{i}^{3}}+3{{i}^{4}}+4{{i}^{5}}.......100{{i}^{101}}
The power of iiin each term increases by 11, because of the same base, which is ii.
Now we got the 2 equations as SiSi and SS. We will be subtracting SiSi from SS which will help us in getting the following result:
⇒S−Si=i+2i2+3i3+4i4.......100i100−(i2+2i3+3i4+4i5.......100i101)\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)
⇒S−Si=i+i2+i3+i4.......i100−100i101\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 ii.
So we will be applying a formula for geometric progression in that Series to find the sum of the series.
⇒S−Si=[i+i2+i3+i4.......i100]−100i101\Rightarrow S-Si=[i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}.......{{i}^{100}}]-100{{i}^{101}}
On calculating the sum we get:
⇒S(1−i)=i(i100−1)i−1−100i101\Rightarrow S\left( 1-i \right)=\dfrac{i\left( {{i}^{100}}-1 \right)}{i-1}-100{{i}^{101}}
⇒S(1−i)=−100i101\Rightarrow S\left( 1-i \right)=-100{{i}^{101}}
⇒S=−100i1011−i\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:
⇒S=−100i101(1+i)(1−i)(1+i)\Rightarrow S=\dfrac{-100{{i}^{101}}\left( 1+i \right)}{\left( 1-i \right)\left( 1+i \right)}
⇒S=−100i101(1+i)(1−i2)\Rightarrow S=\dfrac{-100{{i}^{101}}\left( 1+i \right)}{\left( 1-{{i}^{2}} \right)}
⇒S=−100i101(1+i)(1−(−1))\Rightarrow S=\dfrac{-100{{i}^{101}}\left( 1+i \right)}{\left( 1-(-1) \right)}
⇒S=−100i101(1+i)2\Rightarrow S=\dfrac{-100{{i}^{101}}\left( 1+i \right)}{2}
The power of ii is 101101 which can be written as a sum of 100100 and 11 . Since i100=1{{i}^{100}}=1, so the expression becomes:
⇒S=−50i(1+i)\Rightarrow S=-50i\left( 1+i \right)
⇒S=−50(i+i2)\Rightarrow S=-50\left( i+{{i}^{2}} \right)
⇒S=−50(i−1)\Rightarrow S=-50\left( i-1 \right)
⇒S=50(1−i)\Rightarrow S=50\left( 1-i \right)
∴\therefore The sum of i−2−3i+4......i-2-3i+4......upto 100100 terms, where i=−1i=\sqrt{-1} is 1)50(1−i)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.