Correct option is D. t² is t (t − 1) times greater than t.
Given, t > 1
Let us consider that t²′ is greater than ′ t ′ by ′ n ′
Formally, n + t = t²
⇒n = t² − t
⇒n = t(t − 1)
Therefore, t² is ′ t(t − 1) ′ greater than t.
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