Is That too Much to Calculate?(1)

$(6052+6051+{\dots}+2020+2019+2018)-(2017+2016+{\dots}+5+4+3+2+1) = n^2$

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

$\frac{6052(6052+1)}{2}={\boxed{18316378}}$

$\frac{2017(2017+1)}{2}={\boxed{2035153}}$

$18316378 - 2035153={\boxed{16281225}}$

$16281225=n^2$

$n = {\sqrt{16281225}} ={\boxed{4035}}$

$\color{#3D99F6}{\therefore}$ the answer is $\color{#3D99F6}{\boxed{4035}}$

I have just simply take the arithmetic average in the way from the taken numbers as shown: $n^{2}=(\frac{2018+6052}{2})^{2}={\color{#D61F06}4035}^{2}$ , what is probably the less advanced way to solve it. The red number is our answer.

If you observe carefully the number of terms in every sequence is the square of the resulting number . $1 \quad ({\color{teal}1}~term) = {\color{teal}1}^2$

$2 + 3 + 4 \quad ({\color{#E81990}3}~terms) = {\color{#E81990}3}^2$

$3 + 4 + 5 + 6 + 7 \quad ({\color{#3D99F6}5}~terms) = {\color{#3D99F6}5}^2$

$4 + 5 + 6 + 7 + 8 + 9 + 10 \quad ({\color{#D61F06}7}~terms) = {\color{#D61F06}7}^2$

$\dots \dots$

$2018 + 2019 + 2020 + ......... + 6052 \quad ({\color{#20A900}n}~terms) = {\color{#20A900}n}^2$ Now, we have to find the value of $n$ .

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The numbers $2018, 2019, 2020 , ..... , 6052$ are $\color{#20A900}n$ terms in arithmetic progression with $a = 2018, d = 1$ . The last term in an A.p is given by the formula :

$T_n = a + (n - 1)d$

$\implies 6052 = 2018 + ({\color{#20A900}n} - 1)1 \implies {\color{#20A900}n} + 2017 = 6052$

$\implies \boxed{\color{#20A900}n = 4035}$