Pembahasan Matematika Dasar SIMAK UI 2016 #2

Kali ini saya akan membahas soal Matematika Dasar SIMAK UI 2016. Saya pilihkan yang menarik, karena jarang keluar di SBMPTN atau UM UGM atau soal seleksi yang lain.

Soal

Jika \(1 + \frac{C(2016,1)}{4} + \frac{C(2016,2)}{4^2}+ \frac{C(2016,3)}{4^3} + \ldots + \frac{C(2016,2016)}{4^{2016}} = \left( \frac{5}{4}\right) ^{252n}\), maka nilai \(n=\ldots\)

A. \(2^0\)

B. \(2^1\)

C. \(2^2\)

D. \(2^3\)

E. \(2^4\)

Jawab

Perhatikan bahwa \( \begin{align} (a+b)^n &= \sum_{k=0}^n C(n,k)a^{n-k}b^k\\\\ &= C(n,0)a^{n}b^0 + C(n,1)a^{n-1}b^1 + C(n,2)a^{n-2}b^2 + C(n,3)a^{n-3}b^3 + \ldots + C(n,n)a^{0}b^n \end{align} \)

Jika pada persamaan terakhir kita ganti \(a= 1, b =\dfrac{1}{4}\), dan \(n=2016\),
maka akan diperoleh \( \begin{align} \left(1 + \dfrac{1}{4} \right)^{2016} &= 1 + \dfrac{C(2016,1)}{4} + \dfrac{C(2016,2)}{4^2}+ \dfrac{C(2016,3)}{4^3} + \ldots + \dfrac{C(2016,2016)}{4^{2016}}\\\\ &= \left( \dfrac{5}{4}\right) ^{252n} \end{align} \)

Sehingga \(252n = 2016, n= 8 = 2^3\)

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