prove-by-mathematical-induction-that-4-n-3-n-2-is-a-multiple-of-3-for-all-positive-integral-values-of-n-

Question Number 66102 by Rio Michael last updated on 09/Aug/19

p r o v e b y m a t h e m a t i c a l i n d u c t i o n t h a t 4^{n} + 3^{n} + 2 i s a m u l t i p l e o f 3 f o r a l l

p o s i t i v e i n t e g r a l v a l u e s o f n .

Commented by mathmax by abdo last updated on 09/Aug/19

l e t A_{n} = 4^{n} + 3^{n} + 2

A_{1} = 9 a n d 9 m u l t i p l e o f 3 r e l t i o n t r u e l e t s u p p o s e A_{n} m u l t i p l e

o f 3 \Rightarrow A_{n} = 3 k k i n t e g r o n A_{n + 1} = 4^{n + 1} + 3^{n + 1} + 2

= 4 . (A_{n} - 3^{n} - 2) + 3^{n + 1} + 2 = 4 A_{n} - 4 . 3^{n} - 8 + 3^{n + 1} + 2

= 4 A_{n} - 3^{n} - 6 = 4 (3 k) - 3^{n} - 6 = 3 {4 k - 3^{n - 1} - 2} = 3 k^{^{'}} \Rightarrow

A_{n + 1} i s m u l t i p l e o f 3 t h e r e l a t i o n i s t r u e f o r (n + 1) .

Commented by Prithwish sen last updated on 11/Aug/19

you have to prove it for n = 1 and 2 by putting

n = 1 and n = 2

then assume it is true for n = k

i . e 4^{k} + 3^{k} + 2 is a multiple of 3

now for n = k + 1

4^{k} . 4 + 3^{k} . 3 + 1 = 4 (4^{k} + 3^{k} + 1) - (3^{k} + 3)

∵ this is multiple of 3

∴ it is proved for n = k + 1

Commented by Rio Michael last updated on 09/Aug/19

t h a n k s