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


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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
$let A_n =4^n +3^n +2 A_1 =9 and 9 multiple of 3 reltion true let suppose A_n multiple of 3 ⇒ A_n =3k k integr on A_(n+1) =4^(n+1) +3^(n+1) +2 =4.(A_n −3^n −2)+3^(n+1) +2 =4A_n −4.3^n −8 +3^(n+1) +2 =4 A_n −3^n −6 =4(3k)−3^n −6 =3{4k−3^(n−1) −2}=3k^′ ⇒ A_(n+1) is multiple of 3 the relation is true for (n+1).$
$l e t A_{n} = 4^{n} + 3^{n} + 2$ $A_{1} = 9 a n d 9 m u l t i p l e o f 3 r e l t i o n t r u e l e t s u p p o s e A_{n} m u l t i p l e$ $o f 3 \Rightarrow A_{n} = 3 k k i n t e g r o n A_{n + 1} = 4^{n + 1} + 3^{n + 1} + 2$ $= 4 . (A_{n} - 3^{n} - 2) + 3^{n + 1} + 2 = 4 A_{n} - 4 . 3^{n} - 8 + 3^{n + 1} + 2$ $= 4 A_{n} - 3^{n} - 6 = 4 (3 k) - 3^{n} - 6 = 3 {4 k - 3^{n - 1} - 2} = 3 k^{^{'}} \Rightarrow$ $A_{n + 1} i s m u l t i p l e o f 3 t h e r e l a t i o n i s t r u e f o r (n + 1) .$
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

$t h a n k s$
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