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If Tn = n3 ...

Question

If $T_{n} = n^{3} - 1$ , find the value of 'n' so that $T_{n} = 26$ .

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Solution

We have: $T_{n} = n^{3} - 1$
Hence, $26 = n^{3} - 1, n^{3} = 26 + 1 = 27, n = \sqrt[3]{27} = 3$
$∴ n = 3$

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