To find the sum of first n numbers using the sum of odd numbers and even numbersWe know that the sum of first n numbers is n(n+1)2. So, I tried to derive it using the formula for sum of first n odd numbers and sum of first n even numbers. Here's what I didWhat I tried was to split1+2+3+⋯+ninto(1+3+5+⋯+(2n−1))+(2+4+6+⋯+2n)

Question

To find the sum of first n numbers using the sum of odd numbers and even numbersWe know that the sum of first n numbers is n(n+1)2. So, I tried to derive it using the formula for sum of first n odd numbers and sum of first n even numbers. Here&#039;s what I didWhat I tried was to split1+2+3+⋯+ninto(1+3+5+⋯+(2n−1))+(2+4+6+⋯+2n)

davelucasnp0j · Accepted Answer

Notice that the second part of the expression,&amp;nbsp;(2+4+6+⋯+2n)&amp;nbsp;in particular, is equal to twice the sum of the first n numbers. This is because2+4+6+⋯+(2n−2)+2n2(1)+2(2)+2(3)+⋯+2(n−1)+2(n)2(1+2+3+⋯+(n−1)+n)For you to get the sum of the first n numbers using the formula for odd and even numbers, you have to consider the case where&amp;nbsp;n&amp;nbsp;is odd and&amp;nbsp;n&amp;nbsp;is even. When&amp;nbsp;n&amp;nbsp;is even, then&amp;nbsp;n&amp;nbsp;can be represented as&amp;nbsp;2k&amp;nbsp;where&amp;nbsp;k∈N. Then, there must be&amp;nbsp;k&amp;nbsp;even numbers and&amp;nbsp;k&amp;nbsp;odd numbers. Otherwise, there must be&amp;nbsp;k+1&amp;nbsp;odd numbers and k even numbers.To avoid confusion, we&#039;ll use i as our variable here.We know that1+3+5+⋯+(2i−3)+(2i−1)=i2and&amp;nbsp;2+4+6+⋯+(2i−2)+2i=i(i+1).- Case I: The number of terms is even.Let&amp;nbsp;n&amp;nbsp;be the number of terms. This means that&amp;nbsp;n=2k&amp;nbsp;where&amp;nbsp;k∈N. Hence, we have&amp;nbsp;k&amp;nbsp;even numbers and&amp;nbsp;k&amp;nbsp;odd numbers. Getting the sum of&amp;nbsp;k&amp;nbsp;even numbers and&amp;nbsp;k&amp;nbsp;odd numbers,2+4+6+⋯+2k=k(k+1)1+3+5+⋯+2k−1=k2Adding both the sums,k(k+1)+k2k2+k+k22k2+kk(2k+1)n2(n+1)n(n+1)2- Case II: The number of terms is even.To solve this case, we will use the first case, and just add&amp;nbsp;2k+1. Hence, we have&amp;nbsp;n+1&amp;nbsp;terms. Then,&amp;nbsp;k(k+1)+k2+2k+1k2+k+k2+2k+12k2+3k+1(2k+1)(k+1)(n+1)(n2+1)(n+1)(n+22)(n+1)(n+2)2Our last expression seems to match the formula from the first case, only that&amp;nbsp;n&amp;nbsp;became&amp;nbsp;n+1.

Gonarsu2dw8 · Accepted Answer

In order to find the proof you are searching for you need to emply a trick. As a hint, note that(n+1)2−n2=2n+1

To find the sum of first n numbers

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