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Algebra of Complex Numbers

If Tn = n2 ...

Question

If $T_{n} = n^{2} - 1$ , find (i) $T_{n - 1}$ (ii) $T_{n + 1}$

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Solution

(i) $T_{n - 1}$
Given: $T_{n} = n^{2} - 1$
Put $n = n - 1$
$T_{n - 1} = (n - 1)^{2} - 1$
$= n^{2} - 2 n + 1 - 1$
$= n^{2} - 2 n$
$= n (n - 2)$
$∴ T_{n - 1} = n (n - 2)$

(ii) $T_{n + 1}$
Given: $T_{n} = n^{2} - 1$
Put $n = n + 1$
$T_{n + 1} = (n + 1)^{2} - 1$
$= n^{2} + 2 n + 1 - 1$
$= n^{2} + 2 n$
$= n (n + 2)$
$∴ T_{n + 1} = n (n + 2)$
Hence, $T_{n - 1} = n (n - 2)$ and $T_{n + 1} = n (n + 2)$ .

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(1^{2} - t_{1}) + (2^{2} - t_{2}) + . . . . . . . . . + (n^{2} - t_{n}) = \frac{1}{3} (n^{2} - 1)

t_{n}

t_{n}

is the last term

T_{n} = n^{2} + 1,

S_{2}

(1^{2} - t_{1}) + (2^{2} - t_{2}) + . . . . . + (n^{2} - t_{n}) = \frac{1}{3} n (n^{2} - 1)

t_{n}

(1^{2} - t_{1}) + (2^{2} - t_{2}) + . . . + (n^{2} - t_{n}) = \frac{1}{3} [n (n^{2} - 1)]

, then find the value of

t_{n}

.

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