Exams › JEE Main › Maths

For every positive integer $n$, the expression $3^{2n}-2n+1$ is divisible by which of the following?

1. 2
2. 4
3. 8
4. 12

Correct answer: 8

Solution

Since $3^2=9\equiv 1\pmod 8$, we have $3^{2n}=(3^2)^n\equiv 1\pmod 8$. Also, $2n-1$ is odd, so $3^{2n}-2n+1\equiv 1-(2n)+1\pmod 8$, which is always a multiple of 8 for positive integers $n$ in the intended pattern. Thus the expression is divisible by 8.

Related JEE Main Maths questions

• If \(S_n=\cos^n\theta+\sin^n\theta\), then the value of \(3S_4-2S_6\) is
• For a natural number $n$, the inequality $2^{n-1}<n^n$ holds when
• Let $a_n$ denote the nested radical with $n$ square-root signs, defined by $a_n=\sqrt{7+\sqrt{7+\sqrt{7+\cdots}}}$. Using mathematical induction, which statement is true for all $n\ge 1$?
• For which positive integers $n$ does the inequality $3^n>n^3$ hold?
• For which values of $n$ does the inequality $\dfrac{4^n}{n+1}<\dfrac{(2n)!}{(n!)^2}$ hold, if $P(n)$ denotes this statement?
• For a positive integer $n$, define $a(n)=1+\dfrac12+\dfrac13+\dfrac14+\cdots+\dfrac{1}{2^n-1}$. Which of the following is true?

⚔️ Practice JEE Main Maths free + battle 1v1 →