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Question Number 159488 by Tawa11 last updated on 17/Nov/21

	Answered by mr W last updated on 17/Nov/21

	$T_{n} = 4 T_{n - 1} - 3$ $l e t T_{n} = S_{n} + A$ $S_{n} + A = 4 (S_{n - 1} + A) - 3$ $S_{n} = 4 S_{n - 1} + (3 A - 3)$ $l e t 3 A - 3 = 0, i . e . A = 1$ $S_{n} = 4 S_{n - 1}$ $t h i s i s a G . P .$ $\Rightarrow S_{n} = 4^{n - 2} S_{2}$ $\Rightarrow (T_{n} - 1) = 4^{n - 2} (T_{2} - 1)$ $\Rightarrow T_{n} = 4^{n - 2} (T_{2} - 1) + 1$ $\Rightarrow T_{3} = 4 (T_{2} - 1) + 1 = 4 T_{2} - 3$ $2 T_{3} = 5 T_{2} a s g i v e n$ $2 (4 T_{2} - 3) = 5 T_{2}$ $\Rightarrow T_{2} = 2$ $\Rightarrow T_{n} = 4^{n - 2} (2 - 1) + 1$ $\Rightarrow T_{n} = 4^{n - 2} + 1$ $T_{1} = 4^{- 1} + 1 = \frac{5}{4}$ $T_{2} = 4^{0} + 1 = 2$ $T_{3} = 4 + 1 = 5$ $. . . . . .$
	Commented by Tawa11 last updated on 18/Nov/21

	$God bless you sir . I really appreciate . . .$
	Answered by Rasheed.Sindhi last updated on 18/Nov/21

	$t_{n} = {4 t}_{n - 1} - 3$ $t_{3} = {4 t}_{2} - 3 \land {2 t}_{3} = {5 t}_{2}$ $\Rightarrow 2 ({4 t}_{2} - 3) = {5 t}_{2}$ ${8 t}_{2} - {5 t}_{2} = 6$ $t_{2} = 2$ $t_{3} = 4 (2) - 3 = 5$ $t_{2} = {4 t}_{1} - 3 = 2$ $t_{1} = \frac{5}{4}$ $t_{1}, t_{2}, t_{3} = \frac{5}{4}, 2, 5$
	Commented by Tawa11 last updated on 18/Nov/21

	$God bless you sir, I really appreciate .$
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