The remainder obtained when (2m + 1)^{2n} (m, | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) We need to find the remainder when $(2m + 1)^{2n}$ is divided by $8$. Let $x = 2m + 1$. Since $m \in N$,$x$ is an odd integer. Any odd integer can

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(A) We need to find the remainder when $(2m + 1)^{2n}$ is divided by $8$. Let $x = 2m + 1$. Since $m \in N$,$x$ is an odd integer. Any odd integer can be written as $2m + 1$. Consider $(2m + 1)^2 = 4m^2 + 4m + 1 = 4m(m + 1) + 1$. Since $m(m + 1)$ is the product of two consecutive integers,it is always even. Let $m(m + 1) = 2k$ for some integer $k$. Then $(2m + 1)^2 = 4(2k) + 1 = 8k + 1$. Now,$(2m + 1)^{2n} = ((2m + 1)^2)^n = (8k + 1)^n$. Using the Binomial Theorem,$(8k + 1)^n = \binom{n}{0}(8k)^n + \binom{n}{1}(8k)^{n-1} + \dots + \binom{n}{n-1}(8k) + 1$. All terms except the last one are multiples of $8$. Therefore,$(2m + 1)^{2n} = 8K + 1$ for some integer $K$. Thus,the remainder when divided by $8$ is $1$.

The remainder obtained when $(2m + 1)^{2n}$ $(m, n \in N)$ is divided by $8$ is

Similar Questions

If $(1 + ax)^n = 1 + 8x + 24x^2 + ....,$ then the value of $a$ and $n$ is

If $\{ p \}$ denotes the fractional part of the number $p$,then $\left\{\frac{3^{200}}{8}\right\}$ is equal to

The sum of coefficients in the expansion of $(x + 2y + 3z)^8$ is

$2{C_0} + \frac{2^2}{2}{C_1} + \frac{2^3}{3}{C_2} + \dots + \frac{2^{11}}{11}{C_{10}} = \dots$

The sum of coefficients in $(1 + x - 3x^2)^{2134}$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module