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Elbert

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supaya didapat $x^{50}$, maka harus ada tepat perkalian 50 $x$,, sehingga koefisiennya adalah jumlah semua perkalian 2 bilangan diantara 1 smpai 52...

$1.2 + 1.3 +...+ 1.52 + 2.3 +...+ 2.52 + 3.4 +...+ 3.52 +... 51.52$

ckup ${\frac{(1+2+...52)(1+2+...+52)-({1^2+2^2+3^2+...+52^2})}{2}}$

Edited by Someone
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Meneruskan perhitungan yang diatas ._.

Kita tahu bahwa $\displaystyle \sum_{n=1}^m n^2=\frac{m(m+1)(2m+1)}{6}$.

$=\frac{(1+2+3+...+52)^2-(1^2+2^2+3^2+...+52^2)}{2}$

$ =\frac{\frac{(52 \times 53)^2}{2^2}-\frac{52 \times 53 \times 105}{6}}{2}$

$= \frac{(26 \times 53)^2 - 26 \times 53 \times 35}{2} $

$= \frac{26^2 \times 53^2 - 26 \times 53 \times 35}{2} $

$ =\frac{26 \times 53(26 \times 53 - 35)}{2} $

$ = 689 \times 1325$

Edited by Wildan Bagus W

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