Prove that for any $n \in \mathbb{N}, 2^{n+2} 3^{n}+5n-4$ is divisible by $25$?

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Q Prove that for any $n \in \mathbb{N}, 2^{n+2} 3^{n}+5n-4$ is divisible by $25$?

by using induction

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$2^{n+1+2}3^{n+1}+5(n+1)-4=6\cdot2^{n+2}3^{n}+5n+5-4=(2^{n+2}3^{n}+5n-4)+5(2^{n+2}3^{n}+1)$. So it's enough to show that $2^{n+2}3^{n}+1$ is divisible by $5$. Again induction: $2^{n+1+2}3^{n+1}+1=6\cdot2^{n+2}3^{n}+1=2^{n+2}3^{n}+1+5\cdot(2^{n+2}3^{n}+1) \dots$ — P.., Apr 26 '13 at 07:57

score 4 · Accepted Answer · edited Apr 13 '17 at 12:20

4

Let $f(n)=2^{n+2}3^n+5n-4=4\cdot6^n+5n-4$

$f(m+1)-f(m)=4\cdot6^{m+1}+5(m+1)-4-(4\cdot6^m+5m-4)$ $=5\{5+4(6^m-1)\}$

Now, we know $(6^m-1)$ is divisible by $6-1=5$

$\implies f(m+1)\equiv f(m)\pmod {25}$

Now, $f(1)=2^3\cdot3+5\cdot1-4=25\implies f(1)$ divisible by $25$

Alternatively,

$2^{n+2}3^n+5n-4=4\cdot6^n+5n-4$

$=4(1+5)^n-5n-4$

$=4\left(1+\binom n1 5+\binom n2 5^2+\cdots +\binom n{n-1}5^{n-1}+5^n\right)+5n-4$ (Using Binomial Expansion)

$= 4\{1+5n+5^2\left(\binom n2 +\binom n35 +\binom n45^2+\cdots+\binom n{n-1}5^{n-3}+5^{n-2}\right)\}+5n-4$

As the Binomial coefficients are all integers, all the terms after $\binom n1 5$ is divisible by $5^2=25,$

$5^2\left(\binom n25^2 +\binom n35 +\binom n45^2+\cdots+\binom n{n-1}5^{n-3}+5^{n-2}\right)$ can bewritten as $25k$ where $k$ is an integer

So, $2^{n+2}3^n+5n-4=4\{1+5n+25k\}+5n-4=25(n+4k)$

edited Apr 13 '17 at 12:20

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answered Apr 26 '13 at 06:14

lab bhattacharjee

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can you explain more why 2^n 3^n=6^n and in step 4 how you get 4+4 5n from step 3 I know step 3 is Binomial Theorem, how you get step 3 can you explain more thank you – leena adam Apr 26 '13 at 06:34
@leenaadam, for real $a,b (ab)^n=a^n \cdot b^n$. – lab bhattacharjee Apr 26 '13 at 06:37
how you get 4+4 5n from step 3 I know step 3 is Binomial Theorem, how you get step 4 can you explain more thank you – leena adam Apr 26 '13 at 06:39
Have you noticed the multiplier $4$? Also, $(1+x)^n=1+\binom n1 x+\binom n2 x^2\equiv 1+nx\pmod {x^2}$ for natural $n,x$ – lab bhattacharjee Apr 26 '13 at 06:42
now is clear thanks but last quest where come mode x^2 is these any rule or theorem for mod x^2 this part not clear – leena adam Apr 26 '13 at 06:47
@leenaadam, hope the edited answer helps though the method you are focusing is not based on induction – lab bhattacharjee Apr 26 '13 at 07:42

score 0 · Answer 2 · answered Dec 04 '14 at 01:48

Proof by Induction Second Step observe 2^(n+3)*3(n+1)+5(n+1)-4= 2*2^(n+2)*3*3(n)+5n+5-4= with some algebra 6(2^(n+2)*3(n)+5n-4)-25n+25 Note that by hypothesis 2^(n+2)*3(n)+5n-4 = 25k and therefore the expression can be written as 25(6k-n+1) which is clearly divisible by 25

Prove that for any $n \in \mathbb{N}, 2^{n+2} 3^{n}+5n-4$ is divisible by $25$?

2 Answers2