JEE MAIN - Mathematics (2021 - 26th February Evening Shift - No. 17)
Let $$\alpha$$ and $$\beta$$ be two real numbers such that $$\alpha$$ + $$\beta$$ = 1 and $$\alpha$$$$\beta$$ = $$-$$1. Let pn = ($$\alpha$$)n + ($$\beta$$)n, pn$$-$$1 = 11 and pn+1 = 29 for some integer n $$ \ge $$ 1. Then, the value of p$$_n^2$$ is ___________.
Answer
324
Explanation
Given, $$\alpha$$ + $$\beta$$ = 1, $$\alpha$$$$\beta$$ = $$-$$ 1
$$ \therefore $$ Quadratic equation with roots $$\alpha$$, $$\beta$$ is x2 $$-$$ x $$-$$ 1 = 0
$$ \Rightarrow $$ $$\alpha$$2 = $$\alpha$$ + 1
Multiplying both sides by $$\alpha$$n$$-$$1
$$\alpha$$n$$+$$1 = $$\alpha$$n + $$\alpha$$n$$-$$1 ......(1)
Similarly,
$$\beta$$n + 1 = $$\beta$$n + $$\beta$$n + 1 ..... (2)
Adding (1) & (2)
$${\alpha ^{n + 1}} + {\beta ^{n + 1}} = ({\alpha ^n} + {\beta ^n}) + ({\alpha ^{n - 1}} + {\beta ^{n - 1}})$$
$$ \Rightarrow $$ Pn+1 = Pn + Pn$$-$$1
$$ \Rightarrow $$ 29 = Pn + 11 (Given, Pn + 1 = 29, Pn $$-$$ 1 = 11)
$$ \Rightarrow $$ Pn = 18
$$ \therefore $$ $$P_n^2$$ = 182 = 324
$$ \therefore $$ Quadratic equation with roots $$\alpha$$, $$\beta$$ is x2 $$-$$ x $$-$$ 1 = 0
$$ \Rightarrow $$ $$\alpha$$2 = $$\alpha$$ + 1
Multiplying both sides by $$\alpha$$n$$-$$1
$$\alpha$$n$$+$$1 = $$\alpha$$n + $$\alpha$$n$$-$$1 ......(1)
Similarly,
$$\beta$$n + 1 = $$\beta$$n + $$\beta$$n + 1 ..... (2)
Adding (1) & (2)
$${\alpha ^{n + 1}} + {\beta ^{n + 1}} = ({\alpha ^n} + {\beta ^n}) + ({\alpha ^{n - 1}} + {\beta ^{n - 1}})$$
$$ \Rightarrow $$ Pn+1 = Pn + Pn$$-$$1
$$ \Rightarrow $$ 29 = Pn + 11 (Given, Pn + 1 = 29, Pn $$-$$ 1 = 11)
$$ \Rightarrow $$ Pn = 18
$$ \therefore $$ $$P_n^2$$ = 182 = 324
Comments (0)
