Square peg

Let the Perfect Square be $x^{2}$

Now, since the sum of $n$ numbers are $\frac{n \times (n+1)}{2}$

The average of $n$ will be $\frac{(n+1)}{2}$

So we get,

$\frac{(n+1)}{2} = x^{2}$

or, $n = 2 x^{2} - 1$

For $x=1$ the value of $n$ will be $1$

Since, $n>1$

The next smallest value of $n$ can be found if $x=2$

So, for $x=2$ we get

$n= 2 \times 2^{2} - 1 = 7$

Thus, $\boxed{n=7}$

$1+2+3+4......+n = \frac{n \times (n+1)}{2}$

$\frac{1+2+3+4....+n}{n} = \frac{n+1}{2}$

We will get smallest integer when $\frac{n+1}{2}$ is equal to smallest perfect square greater than 1.

$\frac{n+1}{2} = 4$

$n+1 = 8$

$n = \boxed{7}$

A perfect square is, by definition, an integer that is the square of an integer. The first such number greater than 1 is 4, which can be written as an average of ${1, 2, 3, ..., 7}$ , since $\frac{1+2+3+...+7}{7} = 4$ . Thus, n = $\boxed{7}$ .

The average is just $\dfrac{n+1}{2} = p^2 \implies 2p^2 - 1 = n$ . Now if p is 1, n=0, absurd. If p = 2, then we have $n =\boxed{7}$ .

Lemma: $1+2+3+ ... + n = \frac{n(n+1)}{2}$ for all natural numbers $n$ .

I will prove this by induction. Base case: $n=1$ :

$1 = \frac{1(1+1)}{2}$ . Inductive step: assume the equation holds for $n=k$ for some integer $k$ . Plugging in $k+1$ for $n$ :

$1+2+3+...+k+k+1 = \frac{(k)(k+1)}{2} + k+1$ by the assumption.

$= \frac{(k)(k+1)+2(k+1)}{2}$

$= \frac{(k+1)(k+2)}{2}$ . Thus the equation holds for $n = k+1$ and so the lemma is complete.

The average of the $n$ numbers is thus $\frac{n(n+1)}{2n} = \frac{n+1}{2}$ . Seeing that $n > 4$ , the lowest integer value of $n$ that produces a perfect square is $\boxed{7}$ .

(1 + 2 +...+ n) : n = a^2 ---> ( n(n+1)/2) : n=a^2 ---> n(n+1)/2n = a^2 --->(n+1)/2 = a^2 --->n+1 = 2 . a^2 If a=1, n=1( unsuitable because n>1). If a=2, n=7 ( suitable ). My answer is 7.

SMALLEST N = AVERAGE OF THAT SERIES WHOSE AVERAGE IS SMALLEST PERFECT SQUARE (4) . SO THERE ARE THREE DIGITS BEFORE 4 THEN THERE MUST BE THREE DIGITS AFTER 4 SO ANSWER IS 7

sum of first n natural numbers is n(n+1)/2 so average of these is n(n+1)/2n which is (n+1)/2 so to be a perfect square under root of (n+1)/2 should be a integer so by trial and error 7 is the right answer

(n (n+1))/2n is the Average. This is = ( n+1)/2 The smallest square that bigger than 1 is "4". So (n+1)/2= 4 n+1= 4 2= 8 n=8- 1=7

The answer is 7

1+2+3+.......+n=n(n+1)/2, average of 1,2,3...n=(1+2+3+.......+n)/n=(n+1)/2 ,first perfect square is 4 so n+1/2=4,n=7

We can write the equation

$\frac{1+2+3+4+...+n}{n}=\frac{(n+1)\frac{n}{2}}{n}=\frac{n+1}{2}=x^2$

$x$ is any positive integer.

To find the smallest integer $n$ , we need to minimize the value of $x^2$ . $n>1$ , then $x>1$ .So the minimum value of $x^2=2^2=4$ .

We have the value of $x^2$ , let's solve the equation.

$4=\frac{n+1}{2}$

$n+1=4\times2=8$

$n=8-1=\boxed{7}$ .The smallest integer $n$ is $\boxed{7}$ .

Average of first $n$ integers is sum of them divided by $n$ . Which is...

$\dfrac{1}{n}\sum n = \dfrac{1}{n}\dfrac{n\cdot(n+1)}{2}=\dfrac{n+1}{2}=k^2$

Since, $n>1$ , $k\neq 1$ . So we try the next smallest $k$ which is $2$ . Now we get...

$\dfrac{n+1}{2}=2^2 ~~~ \therefore n=\fbox{7}$

we see that sum of natural numbers upto n is equal to n(n+1)/2 and average is equal to (n(n+1)/2)/n i.e (n+1)/2 .

now (n+1)/2 = k^2 for some k and if we take k=2 then n is smallest which is 7 as n>1

The average of the no.s 1,2,3,4,...n is n(n+1)/2 /n = n+1/2

by this equation as told in question we have to get the perfect square value for average of those numbers,

Now, in (n+1)/2 Change the value of n>1 so that we get a perfect square number.

We get ,

7+1/2 = 4 which is 2^2

Therefore n=7 :)

1+2+3+4........+7=28/7=4 which is square of 2.

2+3+4+n=? then think a number ,when you add it in that equation the sum is a smallest perfect square when n>1

Answer:7