Question

P3 (P+1)
$p3(p + 1) ^{3}$
cube​

Answer 1

Answer:

ANSWER

Given P(1)=1,P(2)=2,P(3)=3,P(4)=5

Let f(x)=(P(x)−x)

f(1)=P(1)−1=1−1=0

f(2)=P(2)−2=2−2=0

f(3)=P(3)−3=3−3=0

∴f(x)=0,x=1,2,3

⇒f(x)=a(x−1)(x−2)(x−3)

P(x)=a(x−1)(x−2)(x−3)+x

Put x=4

5=a(3)(2)(1)+4

⇒a=

6

1

∴P(x)=

6

1

(x−1)(x−2)(x−3)+x

∴P(6)=

6

1

(5)(4)(3)+6=16

Answer 2

Answer:

ans 16 ................think so ...