Permutations & Combinations Question - None

Question

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

Accepted Answer

Step 1: For n = 1, we get $\frac{P^2-(P-1)^1}{P} = \frac{P^2-P+1}{P} = P - 1 + \frac{1}{P}$, which is divisible by P² − P − 1, so result is true for n = 1.

Step 2: Let us assume that the given result is true for n = m ∈ ℕ.

i.e., $P^{m+1}-(P-1)^{2m-1} = k(P^2-P-1)$, where k ∈ ℕ ... (i)

Step 3: Now, $P^{m+2}-(P-1)^{2m+1} = P^{m+2}-(P-1)^2(P-1)^{2m-1}$

$= P^{m+2}-(P-1)^2[k(P^2-P-1)+P^{m+1}]$ [by using Eq. (i)]

$= P^{m+1}[P-(P-1)^2]-(P-1)^2k(P^2-P-1)$

$= P^{m+1}[P-P^2+2P-1]-(P-1)^2k(P^2-P-1)$

$= -P^{m+1}[P^2-P-1]-(P-1)^2k(P^2-P-1)$

$= (P^2-P-1)[k(P-1)^2+P^{m+1}]$

which is divisible by P² − P − 1, so the result is true for n = m + 1.

Therefore, the given result is true for all n ∈ ℕ by induction.

∴ Answer is (c).

Question:

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