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

In the expansion of $(a+1+\frac{1}{a})^n$,where $n \in N$,there are $2029$ terms. Then $n=$

The coefficient of $x^{101}$ in the expression $(5+x)^{500} + x(5+x)^{499} + x^{2}(5+x)^{498} + \ldots + x^{500}$ for $x > 0$ is:

For $|x| < \frac{1}{5}$,the coefficient of $x^3$ in the expansion of $\frac{1}{(1-5 x)(1-4 x)}$ is

Let $r$ be the remainder when $2021^{2020}$ is divided by $2020^2$. Then $r$ lies between

Find $(a+b)^{4}-(a-b)^{4}$. Hence,evaluate $(\sqrt{3}+\sqrt{2})^{4}-(\sqrt{3}-\sqrt{2})^{4}$. (in $\sqrt{6}$)

Vedclass Test Series

Exam Paper Generator

Online Exam Module