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Question: When *P* is a natural number, then \(P^{n + 1} + (P + 1)^{2n - 1}\) is divisible by...

When P is a natural number, then Pn+1+(P+1)2n−1P^{n + 1} + (P + 1)^{2n - 1} is divisible by

A

P

B

P2+PP^{2} + P

C

P2+P+1P^{2} + P + 1

D

P2−1P^{2} - 1

Answer

P2+P+1P^{2} + P + 1

Explanation

Solution

For n =1, we get,

Pn+1+(P+1)2n−1=P2+(P+1)1=P2+P+1P^{n + 1} + (P + 1)^{2n - 1} = P^{2} + (P + 1)^{1} = P^{2} + P + 1,

Which is divisible by P2+P+1P^{2} + P + 1, so result is true for n =1

Let us assume that the given result is true for n=m∈Nn = m \in N

i.e. Pm+1+(P+1)2m−1P^{m + 1} + (P + 1)^{2m - 1} is divisible by P2+P+1P^{2} + P + 1 i.e. pm+1+(p+1)2m−1=k(p2+p+1)p^{m + 1} + (p + 1)^{2m - 1} = k(p^{2} + p + 1) ∀\forall k∈Nk \in N …..(i)

Now. P(m+1)+1+(P+1)2(m+1)−1P ^ { ( m + 1 ) + 1 } + ( P + 1 ) ^ { 2 ( m + 1 ) - 1 }

=pm+2+(p+1)2m+1=pm+2+(p+1)2(p+1)2m−1= p^{m + 2} + (p + 1)^{2m + 1} = p^{m + 2} + (p + 1)^{2}(p + 1)^{2m - 1}

=Pm+2+(P+1)2[k(P2+P+1)−Pm+1]= P^{m + 2} + (P + 1)^{2}\lbrack k(P^{2} + P + 1) - P^{m + 1}\rbrack by using (i)

=Pm+2+(P+1)2⋅k(P2+P+1)−(P+1)2(P)m+1= P^{m + 2} + (P + 1)^{2} \cdot k(P^{2} + P + 1) - (P + 1)^{2}(P)^{m + 1}

=Pm+1[P−(P+1)2]+(P+1)2⋅k(P2+P+1)= P^{m + 1}\lbrack P - (P + 1)^{2}\rbrack + (P + 1)^{2} \cdot k(P^{2} + P + 1)

}{= - P^{m + 1}\lbrack P^{2} + P + 1\rbrack + (P + 1)^{2} \cdot k(P^{2} + P + 1)}$$ $$= (P^{2} + P + 1)\lbrack k \cdot (P + 1)^{2} - P^{m + 1}\rbrack$$ Which is divisible by $p^{2} + p + 1$, so the result is true for $n = m + 1$. Therefore, the given result is true for all $n \in N$ by induction. **Trick :** For *n* = 2, we get, $$\mathbf{P}^{\mathbf{n + 1}}\mathbf{+ (P + 1}\mathbf{)}^{\mathbf{2n - 1}}\mathbf{=}\mathbf{P}^{\mathbf{3}}\mathbf{+ (P + 1}\mathbf{)}^{\mathbf{3}}\mathbf{=}\mathbf{P}^{\mathbf{3}}\mathbf{+}\mathbf{P}^{\mathbf{3}}\mathbf{+ 1 + 3}\mathbf{P}^{\mathbf{2}}\mathbf{+ 3P}\mathbf{ }$$$\mathbf{= 2}\mathbf{P}^{\mathbf{3}}\mathbf{+ 3}\mathbf{P}^{\mathbf{2}}\mathbf{+ 3P + 1}$Which is divisible by $P$. Given result is true for all $n \in N$