If n is natural number, then 3²n -2²n is always divisible by what
Answers
Answer:
Let p(x)=3
2n+2
−8x−9 is divisible by 64 …..(1)
When put n=1,
p(1)=3
4
−8−9=64 which is divisible by 64
Let n=k and we get
p(k)=3
2k+2
−8k−9 is divisible by 64
3
2k+2
−8k−9=64m where m∈N …..(2)
Now we shall prove that p(k+1) is also true
p(k+1)=3
2(k+1)+2
−8(k+1)−9 is divisible by 64.
Now,
p(k+1)=3
2(k+1)+2
−8(k+1)−9=3
2
.3
2k+2
−8k−17
=9.3
2k+2
−8k−17
=9(64m+8k+9)−8k−17
=9.64m+72k+81−8k−17
=9.64m+64k+64
=64(9m+k+1), Which is divisibility by 64
Thus p(k+1)is true whenever p(k) is true.
Hence, by principal mathematical induction,
p(x) is true for all natural number p(x)=3
2n+2
−8x−9 is divisible by 64 n∈N
Step-by-step explanation:
Answer:
\begin{gathered} {9}^{2n} - {3}^{2n} \\ = {( {9}^{n} )}^{2} - {( {3}^{n} )}^{2} \\ = ( {9}^{n} - {3}^{n})( {9}^{n} + {3}^{n})\end{gathered}
9
2n
−3
2n
=(9
n
)
2
−(3
n
)
2
=(9
n
−3
n
)(9
n
+3
n
)
surely for any n, whether even or odd, the factor
( {9}^{n} - {3}^{n}) \:(9
n
−3
n
)
can always be broken in the form:
(9 - 3)(p)(9−3)(p)
where p depends upon n, 9 and 3
so the statement
{9}^{2n} - {3}^{2n} \:9
2n
−3
2n
is always divisible by 9 – 3, i.e., 6.
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