Question

if n+1=2010^2+2011^2, then root 2n+1 is​

Answer 1

$\Huge\rightarrow4021$

$\large\text{\underline{\underline{Question}}}$

If $n+1=2010^{2}+2011^{2}$, find the value of $\sqrt{2n+1}$.

$\large\text{\underline{\underline{Topic}}}$

Polynomial identity

An identity is an equation that is always true. It is known by the following equation.

$\cdots\longrightarrow\boxed{\bold{(a+b)^{2}=a^{2}+2ab+b^{2}}}$

$\large\text{\underline{\underline{Explanation}}}$

$\large\text{[Polynomial idenity]}$

From the following equation -

$\cdots\longrightarrow n+1=2010^{2}+2011^{2}$

we can find $n$ by using a polynomial identity.

$\cdots\longrightarrow n+1=2010^{2}+(2010+1)^{2}$

$\cdots\longrightarrow n+1=2010^{2}+(2010^{2}+2\cdot2010+1^{2})$

On subtraction, -

$\cdots\longrightarrow n=2\cdot2010^{2}+2\cdot2010$

As we want $\sqrt{2n+1}$, for the next step we can multiply both sides by 2.

$\cdots\longrightarrow 2n=4\cdot2010^{2}+4\cdot2010$

Then we can add 1 on both sides.

$\cdots\longrightarrow 2n+1=4\cdot2010^{2}+4\cdot2010+1$

$\cdots\longrightarrow 2n+1=2^{2}\cdot2010^{2}+2\cdot2\cdot2010+1^{2}$

$\large\text{[Polynomial idenity]}$

Now, this is a perfect square. It is observed by a polynomial identity -

$\cdots\longrightarrow\boxed{\bold{(a+b)^{2}=a^{2}+2ab+b^{2}}}$

as known for the perfect square identity.

$\cdots\longrightarrow 2n+1=(2\cdot2010+1)^{2}$

Now, square root gives -

$\text{ \cdots\longrightarrow\sqrt{2n+1}=4021 .}$

$\large\text{[Final answer]}$

So, the required value is $\boxed{4021}$.

Answer 2

$n + 1 = {2010}^{2} + {2011}^{2} \\ n + 1 = {2010}^{2} + {(2010 + 1)}^{2} \\ n + 1 = {2010}^{2} + {2010}^{2} + 2 \times 2010 + 1) \\ n = {2010}^{2} + {2010}^{2} + 2 \times 2010 \\ n = 2 \times {2010}^{2} + 2 \times 2010 \\ 2n = 2(2 \times {2010}^{2} + 2 \times 2010) \\ 2n = 4 \times {2010}^{2} + 4 \times 2010 \\ 2n + 1 = 4 \times {2010}^{2} + 4 \times 2010 + 1 \\ 2n + 1 = (2 \times {2010}^{2} + 2 \times 2 \times 2010 \times 1 \\ 2n + 1 = (2 \times 2010 { +1)}^{2} \\ \sqrt{2n + 1} = \sqrt{(2 \times 2010 { + 1}^{2} )} \\ \sqrt{2n + 1} = 2 \times 2010 + 1 \\ \sqrt{2n + 1} = 4020 + 1 = 4021 \\ \huge \mathfrak \gray { Thank\:You}$