Preparing these solved questions will help with homework and exams as well.Īn odd number is a whole number that cannot give two equivalent resultants when divided by Two. Sum of Odd Natural Numbers is a great chapter explaining the basics of graphs. Sum of Odd Natural Numbers Formula will make the subject interesting and fun to learn. The solution provided will help to solve any questions related to this chapter. The formula provides for easy exam preparation and completing homework. These explanations are easy to understand and clear all doubts at once. The Sum of Odd Natural Numbers Formula is explained in simple steps here. Short Description of Sum of Odd Natural Numbers These solutions will help understand the chapter both critically and logically. During the exams, if you follow these solved questions, it will clear all doubts. Sum of Odd Numbers Formula is an easy and scoring chapter. These solved questions of the Sum of Odd Natural Numbers Formula will make revision easier for students before the exams. These solutions contain essential question answers and solvable questions that may come in the examination. These solutions will widely help during the exams and also for its preparation. I didn't think it was easy to find a solid proof with either of them, so I'd be interested to hear from you if anyone makes any progess.Īnother obvious way of extending the problem is to try considering the number of ways of expressing a number as the sum of four odd numbers, or five or more.Sum of Odd Natural Numbers Formula will give you a vivid description of this chapter on odd numbers. You might find it easier to work with this version of the problem in your attempt to prove classĢYP's formula. The equation $a + b + c = n$ (for odd $a,b,c,n$) has as many solutions as this equation: $$(a+1) + (b+1) + (c+1) = (n+3).$$ Each of the bracketed quantities is even, so dividing through by a factor of 2 gives $$x + y + z = t $$ where $x,y,z$ are any integers, and $t$ is any integer greater then 2. If you thought that the pattern forming above looked suspiciously similar to that which we saw previously for only odd numbers, you would be correct, as I am about to demonstrate. The table above shows the number of ways of summing any 3 numbers to achieve the required total $t$. This will be the number of solutions to the problem because for each pair $\ Now all that remains is to count the number of possibilities for $a$ for each possible value of $c$. The smallest possible value of $a$ is $a= n-2c$ (which occurs when $n-2c> 0$ and $b=c$) and the largest is $a=(n-c)/2$ (which occurs when $a=b$). Once we've decided the range of possibilities for $c$, the possibilities for $a$ can also be limited. This restricts the range of values of $c$ to the interval $n/3 \leq c \leq n-2$. This is evident when you consider that $c$ is defined to be the largest of $a,b,c$ and that they must sum to $n$. Now that we've done that, we can also say that $c$ is at least the smallest odd integer greater than or equal to $n/3$ (where n is the required total). This way none of the solutions are repeated by having the same numbers in a different order. First of all, you should start with a trick which often comes in handy if you are trying to find a certain number of solutions and the order doesn't matter, that is to label them $a,b,c$ and define To do this you would need to consider how to limit your search. If you want to solve the problem differently, you might be interested in programming a computer to find the number of solutions for you. Will probably be much more difficult to show conclusively that their result concerning 3 odd numbers is correct. For each of the $k$ odd numbers there will be another such that the sum of the two is $n$ and the two cases occur according to whether $k$ is even or odd. If the even number $n$ is equal to $2k$, then the number of odd numbers less than $n$ is $k$. To start with, class 2YP found that $P_2(n) = n/4$ when $n$ is divisible by 4 and $P_2(n) = (n+2)/4$ when $n$ is an even number not divisible by 4. I'd like to introduce the following notation: let $P_x(n)$ be the number of ways in which $n$ can be expressed as the sum of $x$ odd numbers where we only count each set of $x$ numbers once, that is we ignore the order in which the numbers occur. They investigated the number of ways of expressing an integer as the sum of odd numbers. You may have seen the solution by Class 2YP from Madras College to a problem which they were inspired to consider after working on the problem called Score from the June Six.
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