How to solve 1+2+3+4+5+6+7…..45

Welcome to my article how to solve 1+2+3+4+5+6+7…..45. This question is taken from the simplification lesson.
The solution of this question has been explained in a very simple way by a well-known teacher by doing addition, subtraction, and fractions.
For complete information on how to solve this question how to solve 1+2+3+4+5+6+7…..45, read and understand it carefully till the end.

Let us know how to solve this question how to solve 1+2+3+4+5+6+7…..45

First write the question on the page of the notebook.

how to solve 1+2+3+4+5+6+7…..45

To solve this question, first we will write it in this way.

\displaystyle \Rightarrow \mathbf{1}+2+\mathbf{3}+4+\mathbf{5}+6+\mathbf{7}+………+44+45

We have to find the sum of this question “1+2+3+4+………………………+45”

By looking at this question, we get to know that,

We have to find the sum of the first 45 natural numbers.

now,

we know that,

To find the sum of n natural numbers, we use this formula –

Sum of n natural numbers = \displaystyle \text{ }\frac{{\left[ {\mathbf{n}\left( {\mathbf{n}+\mathbf{1}} \right)} \right]}}{\mathbf{2}}

Here,

n = last number = 45

now,

{sum of n natural numbers = \displaystyle \frac{{\left[ {n\left( {n+1} \right)} \right]}}{2}}

Substituting the value of n = 45 in this formula,

we find,

Sum = \displaystyle \frac{{\left[ {45\text{ }\times \text{ }\left( {45+1} \right)} \right]}}{2}

= \displaystyle \frac{{\left[ {45\text{ }\times \text{ }46} \right]}}{2}

= 45 × 23

= 1035

Hence,

1+2+3+4+…………+45 = 1035 [ Answer ]

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