The sequence about the year

The formula for the sum of the first $n$ positive perfect squares is

$\frac{n(n+1)(2n+1)}{6}.$

The formula for the sum of the first $n$ positive integers is

$\frac{n(n+1)}{2}.$

Thus, by dividing the two quantities, we get

$\frac{\frac{n(n+1)(2n+1)}{6}}{\frac{n(n+1)}{2}} = \frac{2n+1}{3}.$

Substituting, we get

$\frac{4029}{3}=\boxed{1343}.$

For those who dont have the algebraic background to solve this, this is an alternative solution: ${ T }_{ 1 }=\frac { { 1 }^{ 2 } }{ 1 } \\ { T }_{ 2 }=\frac { { 1 }^{ 2 }+{ 2 }^{ 2 } }{ 1+2 } \\ { T }_{ 3 }=\frac { { 1 }^{ 2 }+{ 2 }^{ 2 }+{ 3 }^{ 2 } }{ 1+2+3 } \\ ...\\ { T }_{ 1 },\quad { T }_{ 2 },\quad { T }_{ 3 },\quad { T }_{ 4 }...\\ 1,\frac { 5 }{ 3 } ,\frac { 7 }{ 3 } ,3...\\ \frac { 3 }{ 3 } ,\frac { 5 }{ 3 } ,\frac { 7 }{ 3 } ,\frac { 9 }{ 3 } ...\\ \therefore { T }_{ n }=\frac { 2n+1 }{ 3 } \\ { T }_{ 2014 }=\frac { 2(2014)+1 }{ 3 } =1343$

$We\quad know\quad that\quad :\\ 1²\quad +\quad 2²\quad +\quad 3²\quad +...+n²\quad =\quad \frac { n(n+1)(2n+1) }{ 6 } \\ and\quad :\\ 1\quad +\quad 2\quad +\quad 3\quad +\quad ...+\quad n\quad =\quad \frac { n(n+1) }{ 2 } \\ So\quad :\\ \frac { 1²+2²+3²+...+n² }{ 1+2+3+...+n } =\quad \frac { \frac { n(n+1)(2n+1) }{ 6 } }{ \frac { n(n+1) }{ 2 } } \quad =\quad \frac { \\ \frac { n(n+1)(2n+1) }{ 3 } }{ n(n+1) } \quad =\quad \frac { 2n+1 }{ 3 } \\ \\ For\quad n\quad =\quad 2014\quad we\quad have\quad :\\ \\ \frac { 2n+1 }{ 3 } =\quad \frac { 4029 }{ 3 } =\quad 1343\\ \\ \therefore \quad \frac { 1²+2²+3²+...+2014² }{ 1+2+3+...+2014 } \quad =\quad 1343$

We have $(n+1)^3-n^3=3*n^2+3*n+1$ Then, let $A= \frac{1^2+2^2+3^2+\ldots+2012^2+2013^2+2014^2}{1+2+3+\ldots+2012+2013+2014}$ Then $3*(A+1)= \frac{2^3-1^3+3^3-2^3+4^3-3^3+\ldots+2013^3-2012^3+2014^3-2013^3+2015^3-2014^3-2014}{1+2+3+\ldots+2012+2013+2014}$

After cancellation we have $3*(A+1)= \frac{2015^3-2015}{1+2+3+\ldots+2012+2013+2014}$ After applying the formula of the sum of the first $n$ positive integers we have $3*(A+1)= \frac{2015^3-2015}{2015*2014*\frac{1}{2}}$ Then we have $3*(A+1)=4032$ So we get the value of A $A=1343$

if we calculate the first terms as : $\frac{1^{2}}{1}$ = 1 = $\frac{3}{3}$ and the second term can be written as : $\frac{1^{2}+2^{2}}{1+2}$ = $\frac{5}{3}$ .......... so in general : the $n^{th}$ term can be represented as : $\frac{1+2n}{3}$ .......... so : the $2014^{th}$ term is $\frac{1+2014*2}{3}$ = 1343

Okay if you observe the series and divide it .. 1^2÷1=1 i.e. 3/3 (1^2+2^2)÷(1+2)=5/3...... .... And so on We have to find series till 2014 so we will use the 2015th odd number in numerator and 3 in denominator Which gives us 4029/3=1343 Which is the answer

the hint made the sum much more easier. The sum should have been given without a hint. Then it would be more challenging

The solution is 1343.

Python:
a=0
b=0
c=1
while(c<=2014):
a+=c*c
b+=c
c+=1
print(a/b)

screwmath

= $\frac{2014 X 2015 X 2 X 2014 +1}{2 X 3 X (1+2+3+...+2014) }$ = $\frac{2014 X 2015 X 4029}{2 X 3 X 2015 X 1007 }$ = $\frac{2 X 1 X 4029}{2 X 3 X 1 X 1}$ = $\frac{4029}{3}$ = 1343

$\Large{\frac{1^2 + 2^2 + 3^2 +......+ 2012^2 + 2013^2 + 2014^2}{1 + 2+ 3 + ......+ 2012 + 2013 + 2014}}$

$\Leftrightarrow$ $\Large{\frac{\frac{1}{6}n(n + 1)(2n + 2)}{\frac{1}{2}n (1 + Un)}}$

$\Leftrightarrow$ $\Large{\frac{\frac{1}{6}n(n + 1)(2n + 2)}{\frac{1}{2}n (1 + n)}}$ $, for \quad Un = n$

$\Leftrightarrow$ $\frac{1}{3} \times (2n + 2)$

$\Leftrightarrow$ $\frac{1}{3} \times (2 \times 2014 + 2) = \boxed{1343}$

most people uses a direct formula for this, but i found that sometimes it is easier to do a direct approach to these kind of questions e.g. 1^2/1 = 1 = 3/3

(1^2+2^2)/(1+2) = 5/3

(1^2+2^2+3^2)/(1+2+3) = 14/6 = 7/3

(1^2+2^2+3^2+4^2)/(1+2+3+4) = 30/10 = 3 = 9/3

from these results we can see a pattern

the denominator will always be 3, while the numerator is form in a pattern; 3, 5, 7, 9, ...

so the only thing need to be done is find the pattern equation for the numerator.

for this example, the pattern can be simplify;

3, 5, 7, 9, ... = 3, (3+2), (3+4), (3+6), ...

from that, an equation to the pattern is made: 3+2(n-1).

now we need to test if the equation above will satisfy the pattern.

Testing

n | result

1 | 3

2 | 5

3 | 7

4 | 9

the equation satisfy the pattern!

therefore, the complete equation to the example is (3+2(n-1))/3

now, since the queation ends with 2014, we subtitute n as 2014.

(3+2(2014-1))/3 = 4029/3 = 1343