A number theory problem by SARAN .P.S

Number Theory Level 1

If the sum of the first n positive integers is equal to 55, then what is the sum of the first n square numbers?

2255 385 325 1025

3 solutions

Dawar Husain
Dec 20, 2014

The problem is not correctly stated. At first, I thought it asked sum of nos. till $n^2$ .
$\dfrac{n(n+1)}{2}\ =\ 55$

$n^2 + n-110=\ 0$

$(n-10)\ (n+11)=0$

$\implies n=10\ or\ -11$

As $n$ can't be -ve, we take $n=10$ . Now, we have to find the sum of squares.

$=\ \dfrac{n(n+1)(2n+1)}{6}=\dfrac{10\times 11 \times 21}{6}=\ \boxed{385}$

Thanks, I have updated the phrasing accordingly.

In future, if you spot any errors with a problem, you can “report” it by selecting the “dot dot dot” menu in the lower right corner. You will get a more timely response that way.

Calvin Lin Staff - 6 years, 5 months ago

Thank you! I am new here so I got confused. I reported this question. https://brilliant.org/problems/challenging-buoyancy/?group=p5mPnHlX0dzj&ref_id=534438

Dawar Husain - 6 years, 5 months ago

Thnx, I have the same answer as you

Nam Cao Vũ Hoàng - 6 years, 5 months ago

Amresh Giri
Dec 22, 2014

I agree ! The question was a little bit confusing ! But it was very easy anyway :-)

Jitendra Sharma
Dec 27, 2014

It should have been asked for sum of the squares of first n positive integers

A number theory problem by SARAN .P.S

3 solutions

0 pending reports