prove-by-mathematical-induction-7-n-3n-4-4-n-1-divided-by-9-

Question Number 111132 by bemath last updated on 02/Sep/20

prove by mathematical induction

\Rightarrow 7^{n} - (3 n + 4) \times 4^{n - 1} divided by 9

Answered by john santu last updated on 02/Sep/20

l e t p (n) = 7^{n} - (3 n + 4) . 4^{n - 1}

(1) p (1) = 7 - (3.1 + 4) . 4^{0} = 7 - 7 = 0 (t r u e)

(2) a s s u m e f o r n = k \to p (k) d i v i s i b l e b y 9

w e h a v e p (k) = 7^{k} - (3 k + 4) . 4^{k - 1} \equiv 9 m, m \in Z

= 7^{k} - 3 k . 4^{k - 1} - 4^{k} \equiv 9 m

= 7^{k} - 4^{k} - 3 k . 4^{k - 1} \equiv 9 m

(3) w e w a n t t o p r o v e f o r n = k + 1

p (k + 1) i s d i v i s i b l e b y 9

p (k + 1) = 7^{k + 1} - (3 (k + 1) + 4) . 4^{k}

\Leftrightarrow 7 . 7^{k} - (3 k + 4 + 3) . 4^{k}

\Leftrightarrow 7 . 7^{k} - (3 k + 7) . 4^{k}

\Rightarrow 7 . 7^{k} - 3 k . 4^{k} - 7 . 4^{k}

\Rightarrow 7 (7^{k} - 4^{k}) - 3 k . 4^{k}

\Rightarrow 7 (7^{k} - 4^{k} - 3 k . 4^{k - 1}) + 21 k . 4^{k - 1} - 3 k . 4^{k}

\Rightarrow 7 (9 m) + 3 k . 4^{k - 1} (7 - 4)

\Rightarrow 63 m + 9 k . 4^{k - 1}

\Rightarrow 9 (7 m + k . 4^{k - 1}) i s d i v i s i b l e b y 9 .

q . e . d