Welcome to my article i+i^2+i^3+i^4 upto 99 terms. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question i+i^2+i^3+i^4 upto 99 terms, read and understand it carefully till the end.

Let us know how to solve this question i+i^2+i^3+i^4 upto 99 terms.

First write the question on the page of the notebook.

How to solve i+i^2+i^3+i^4 upto 99 terms

First write this question as follows and then solve,

\displaystyle i+{{i}^{2}}+{{i}^{3}}+{{i}^{4}}……………99terms

here,

a = i

n = 9

r = \displaystyle \frac{{{{i}^{2}}}}{i} = i

Formula

\displaystyle {{S}_{n}}=\frac{{a({{r}^{n}}-1)}}{{r-1}}

Substituting the values of a , r , n in this formula,

\displaystyle {{S}_{n}}=\frac{{i({{i}^{{99}}}-1)}}{{i-1}}

\displaystyle {{S}_{n}}=\frac{{i({{i}^{{96}}}\times {{i}^{3}}-1)}}{{i-1}}

we know that ,

\displaystyle i=i

\displaystyle {{i}^{2}}=-1

\displaystyle {{i}^{3}}=-i

\displaystyle {{i}^{4}}=1

so,

\displaystyle {{S}_{n}}=\frac{{i\left[ {{{{{{{{(i)}}^{4}}}}}^{{24}}}\times {{i}^{3}}-1)} \right]}}{{i-1}}

\displaystyle {{S}_{n}}=\frac{{i\left[ {{{1}^{{24}}}\times -i-1)} \right]}}{{i-1}}

\displaystyle {{S}_{n}}=\frac{{i\left[ {1\times -i-1)} \right]}}{{i-1}}

\displaystyle {{S}_{n}}=\frac{{-{{i}^{2}}-i}}{{i-1}}

\displaystyle {{S}_{n}}=\frac{{-{{{(i)}}^{2}}-i}}{{i-1}}

\displaystyle {{S}_{n}}=\frac{{-(-1)-i}}{{i-1}}

\displaystyle {{S}_{n}}=\frac{{1-i}}{{i-1}}

\displaystyle {{S}_{n}}=\frac{{-1(i-1)}}{{i-1}}

\displaystyle {{S}_{n}}=-1

so,

i+i^2+i^3+i^4 upto 99 terms = -1 [Answer]

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