Assuming \((X - I)^{-1}\) exists, what will be \(I + X + X^{2} + \ldots + X^{n-1}\) ?
MH SET - 2025, SET Exam - 2025,
Answer: C
The sum is \((X^{n} - I)(X - I)^{-1}\).
Multiply the sum by \(X - I\) and the terms telescope:
\[ \begin{aligned} &(I + X + \ldots + X^{n-1})(X - I) \\ &= (X + X^{2} + \ldots + X^{n}) - (I + X + \ldots + X^{n-1}) \\ &= X^{n} - I \end{aligned} \]
Multiplying on the right by \((X - I)^{-1}\) gives the result; all the matrices are polynomials in X, so they commute and the order does not matter.
Option B has the sign reversed: \((I - X^{n})(X - I)^{-1}\) is the negative of the sum. Options A and D raise \(I - X\) or \(X - I\) to the nth power, which is not the same as \(X^{n} - I\).
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