Sum that requires no adding

Number Theory Level 2

Consider the table:

$\begin{array} {c} 1=1 \\ 3+5= 8 \\ 7+9+11= 27 \\ 13+15+17+19= 64 \end{array}$

What is the sum of the numbers in the tenth row ( that should be indicated on the right hand side)?

2000 2500 1000 1005 1500

3 solutions

Peter van der Linden
Apr 19, 2017

Every nth row adds the next n odd numbers. Since we are looking for the 10th row, we are adding the 46th till 55th odd number. The sum of this is: $55^2 - 45^2 = 1000$

Nice idea and solution.

Hana Wehbi - 4 years, 1 month ago

Tnx. I saw the 3rd power too, but wanted to post another idea.

Peter van der Linden - 4 years, 1 month ago

No problem.

Hana Wehbi - 4 years, 1 month ago

Can you find a general term for this pattern that can be used so that we don't have to methodically work upwards to find the next $n$ odd numbers for the $n^{th}$ row??

Zach Abueg - 4 years, 1 month ago

Excuse the double ??, sorry about that.

Zach Abueg - 4 years, 1 month ago

Well it can be done, I used the sumformula for integers to find the 46th till 55th odd numbers. This formula is known as 0,5n(n+1). So for the nth row you use 0,5n(n+1) - 0,5(n - 1)n

Peter van der Linden - 4 years, 1 month ago

Yes, we can generalize this. I will post the proof later.

Hana Wehbi - 4 years, 1 month ago

Hana Wehbi
Apr 19, 2017

Hint: note that the right hand side is $n^3$ .

A Former Brilliant Member
Apr 21, 2017

$1^2(1)=1$

$2^2(2)=8$

$3^2(3)=27$

$4^2(4)=64$

It follows that in the tenth row,

$10^2(10)=1000$

Nice thought.

Hana Wehbi - 4 years, 1 month ago

Sum that requires no adding

3 solutions

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