Statement -1: For every natural number n,(n + | vedclass

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(D) For Statement $-2$: By Fermat's Little Theorem,for any prime $p$ and integer $n$,$n^p \equiv n \pmod{p}$. Here $p = 7$,so $n^7 \equiv n \pmod{7}$,

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Statement $-1$ is true,Statement $-2$ is true; Statement $-2$ is a correct explanation for Statement $-1$.

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(D) For Statement $-2$: By Fermat's Little Theorem,for any prime $p$ and integer $n$,$n^p \equiv n \pmod{p}$. Here $p = 7$,so $n^7 \equiv n \pmod{7}$,which means $n^7 - n$ is divisible by $7$. Thus,Statement $-2$ is true.For Statement $-1$: Expand $(n + 1)^7$ using the Binomial Theorem: $(n + 1)^7 = n^7 + 7n^6 + 21n^5 + 35n^4 + 35n^3 + 21n^2 + 7n + 1$. Then $(n + 1)^7 - n^7 - 1 = 7n^6 + 21n^5 + 35n^4 + 35n^3 + 21n^2 + 7n = 7(n^6 + 3n^5 + 5n^4 + 5n^3 + 3n^2 + n)$. This expression is clearly divisible by $7$. Thus,Statement $-1$ is true.Note that Statement $-1$ can be written as $(n+1)^7 - (n+1) - (n^7 - n) = 7k$. Since both $(n+1)^7 - (n+1)$ and $n^7 - n$ are divisible by $7$ (by Statement $-2$),their difference is also divisible by $7$. Therefore,Statement $-2$ is a correct explanation for Statement $-1$.

Statement $-1$: For every natural number $n$,$(n + 1)^7 - n^7 - 1$ is divisible by $7$.
Statement $-2$: For every natural number $n$,$n^7 - n$ is divisible by $7$.

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Statement $-1$: For every natural number $n$,$(n + 1)^7 - n^7 - 1$ is divisible by $7$. Statement $-2$: For every natural number $n$,$n^7 - n$ is divisible by $7$.