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GATE Overflow Test Series | Discrete Mathematics | Test 3 | Question: 11

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Solve by substitution,

$x_n = 5x_{n-1} + 3$

$\qquad = 5(5.x_{n-2}+3) + 3$

$\qquad = 5^2(x_{n-2}) + 5.3 + 5^0.3$

$\qquad = \ldots$

$\qquad = 5^{n-1}(x_{n-(n-1)}) + \underbrace{(5^{n-2}.3 + 5^{n-1}.3 + \ldots +  5^0.3)}_{\text{Sum to n-1 terms of GP with r = 5, a = 3}}$

$\qquad = 5^{n-1}(x_1) + 3.\dfrac{5^{n-1}-1}{4}$

$\qquad = \dfrac{5.3.5^{n-1} - 3}{4}$

$\qquad = 3.\dfrac{5^{n} - 1}{4}$

$\implies x_5 = 3.\dfrac{(5^5-1)}{4} = 2343.$
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