prove that 2n+3 is odd positive integer
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Step-by-step explanation:
Any odd positive integer can be written as n=2k+1
n2=4k2+4k+1=4k(k+1)+1
Now, for two consecutive integers, there exists one integer which is even.
I think you are done.
ANSWER
Let, P(n)=a
n
+b
n
P(1)=a+b, which is divisible by a+b
Now let P(k)=a
k
+b
k
is divisible by a+b, where k is an odd integer.
⇒a
k
+b
k
=(a+b)f(a,b)⇒(1)
Now, P(k+2)=a
k+2
+b
k+2
=a
2
[(a+b)f(a,b)−b
k
]+b
k+2
=a
2
f(a,b)(a+b)−a
2
b
k
+b
k+2
=a
2
f(a,b)(a+b)−b
k
(a
2
−b
2
)
=(a+b)[a
2
f(a,b)−b
k
(a−b)], which is divisible by (a+b)
Hence a
n
+b
n
is divisible by (a+b) for all odd positive integral n
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