Question

Find the sum of natural numbers from 100 400​

Answer 1

$\rm 75250$

$\Large\textrm{Explanation}$

Let us consider the two sequences. One is the first 99 natural numbers, and the other is the first 400 natural numbers.

Both series contain numbers from 1 to 99. Then how do we eliminate them?

We can eliminate them by subtraction. Think about the way to solve system equations.

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So, let us show the series of the first $\rm n$ terms by $\rm S_{n}$. What is the formula?

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It is easy to derive the formula.$\;$$\;$

We have,

$\boxed{\begin{aligned}\rm S_{n}&\rm=1+2+3+\cdots+(n-2)+(n-1)+n\\\\\rm S_{n}&\rm=n+(n-1)+(n-2)+\cdots3+2+1\end{aligned}}$

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We can find it by reversing the order of calculation.

$\rm2S_{n}=n(n+1)$

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Hence,

$\rm\therefore S_{n}=\dfrac{n(n+1)}{2}$

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We showed the main idea of the formula!

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Let's apply the formula and get the result.

$\rm S_{400}-S_{99}$

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$\rm=\dfrac{400\cdot401}{2}-\dfrac{99\cdot100}{2}$

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$\rm=80200-4950$

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$\rm=75250$

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$\Large\textrm{Learn More}$

$\textbf{- Term Explanation}$

Series: A sum of the numbers.

Sequence: A set of numbers defined mathematically.

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$\textbf{- The Sum of the Perfect Squares and Cubes}$

For perfect squares,

$\boxed{\rm S_{n}=\dfrac{n(n+1)(2n+1)}{6}}$

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For perfect cubes,

$\boxed{\rm S_{n}=\left\{\dfrac{n(n+1)}{2}\right\}^{2}}$