Tolong dibantu dong hari ini juga please

Tolong dibantu dong hari ini juga please

$10 sqrt{10} . sqrt[4]{10} . sqrt[8]{10}. sqrt[16]{10} ...... sqrt[ {2}^{n} ]{10} \ \ = {10}^{1} times {10}^{ frac{1}{2} } times {10}^{ frac{1}{4} } times {10}^{ frac{1}{8} } .... {10}^{ frac{1}{ {2}^{n} } } \ \ = 1 + frac{1}{2} + frac{1}{4} + .... \ \ = 1 + frac{3}{4} + frac{1}{8} + ... \ \ = 1 + frac{7}{8} + frac{1}{16} + .... \ \ = 1 + frac{15}{16} + ... \ \ = 2 \ \ \ jadi \ \ {10}^{2} \ \ = 10 times 10 \ \ = 100$

Jadi, Jawabannya adalah B

Semoga Membantu

Jawab:

Penjelasan dengan langkah-langkah:

$lim_{n to infty} 10sqrt{10}sqrt[4]{10}sqrt[8]{10}...sqrt[2^n]{10}\\=lim_{n to infty} 10.10^{frac{1}{2}}.10^{frac{1}{4}}.10^{frac{1}{8}}...10^{frac{1}{2^n}}\\=lim_{n to infty} 10^{1+frac{1}{2}+frac{1}{4}+frac{1}{8}+...+frac{1}{2^n}}\\\mencari~jumlah~frac{1}{2}+frac{1}{4}+frac{1}{8}+...+frac{1}{2^n}\\misal\\S_n=frac{1}{2}+frac{1}{4}+frac{1}{8}+...+frac{1}{2^{n-1}}+frac{1}{2^n}~~~~~...kedua~ruas~dikali~2\\2S_n=frac{2}{2}+frac{2}{4}+frac{2}{8}+...+frac{2}{2^n}$

$2S_n=1+frac{1}{2}+frac{1}{4}+frac{1}{8}+...+frac{1}{2^{n-1}}\\2S_n=1+(S_n-frac{1}{2^n})\\2S_n-S_n=1-frac{1}{2^n}\\S_n=1-frac{1}{2^n}\\\kembali~ke~soal\\=lim_{n to infty} 10^{1+frac{1}{2}+frac{1}{4}+frac{1}{8}+...+frac{1}{2^n}}\\=lim_{n to infty} 10^{1+1-frac{1}{2^n}}\\=lim_{n to infty} 10^{2-frac{1}{2^n}}\\=10^{(lim_{n to infty} 2-frac{1}{2^n})}\\=10^{(2-0)}\\=10^2\\=100$

#sejutapohon

Mapel: Matematika

Kelas : 11

Bab : Limit

Kata Kunci : limit, tak hingga

Kode Kategorisasi: 11.2.8