prove that 2*7^n-3*5^n+1 is divisible by 24
Answers
Answer:
Let P(n):2.7
n
+3.5
n
−5 is divisible by 24
We note that P(n) is true when n=1, since 2.7+3.5−5=24. which is divisible by 24.
Assume that P(k) is true.
i.e. 2.7
k
+3.5
k
−5=24q when q∈N -------------- ( 1 )
Now, we have to prove that P(k+1) is true whenever P(k) is true.
We have
2.7
k+1
+3.5
k+1
−5
⇒ 2.7
k
.7
1
+3.5
k
.5
1
−5
⇒ 7[2.7
k
+3.5
k
−5−3.5
k
+5]+3.5
k
.5−5
⇒ 7[24q−3.5
k
+5]+15.5
k
−5
⇒ 2×24q−21.5
k
+35+15.5
k
−5
⇒ 7×24q−6.5
k
+30
⇒ 7×24q−6(5
k
−5)
⇒ 7×24q−6(4p) [ (5
k
−5) is multiple of 4 ]
⇒ 7×24q−24p
⇒ 24(7p−q)
⇒ 24×r;r=7p−q. is some natural number ---------- ( 2 )
The expression on the R.H.S oof ( 1 ) is divisible by 24. Thus P(k+1) is true whenever P(k) is true.
Hence, by principle of mathematical induction , P(n) is true for all n∈
Step-by-step explanation:
So, for n=1, given expession is divisible by 24. Let for any k∈N, given expression is divisible by 24. Then, 2⋅7k+3⋅5k-5=24c, where c is a natural number.