Blog Competition 2019 #2019GANTIBIMBEL

Buktikan bahwa \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{n}{2^n} = 2 - \frac{n + 2}{2^n}

Buktikan bahwa \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{n}{2^n} = 2 - \frac{n + 2}{2^n}
Pembahasan :
  • Langkah 1
 \frac{1}{2} = 2 - \frac{(1)+2}{2^1} = 2 - \frac{3}{2}
\frac{1}{2} = \frac{1}{2}      (terbukti)
  • Langkah 2 (n = k)
\frac{1}{2} + \frac{2}{2^2} + \cdots + \frac{2}{2^k} = 2 - \frac{k + 2}{2^k}
  • Langkah 3 (n = k + 1)
\frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{k}{2^k} + \frac{k + 1}{2^{k + 1}} = 2 - \frac{k + 3}{2 ^{k +1}}
Dibuktikan dengan:
 = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \cdots + \frac{k}{2^k} + \frac{k + 1}{2^{k + 1}} = 2 - \frac{k + 2}{2^k} + \frac{k + 1}{2^{k + 1}}     (kedua ruas dikali \frac{k+1}{2^{k+1}})
 = 2 - \frac{2(k + 2)}{2^{(k + 1)}} + \frac{k + 1}{2^{k +1}}      (2k dimodifikasi menjadi 2k+1)
= 2 -\frac{2k + 4}{2^{(k + 1)}} + \frac{k + 1}{2^{k + 1}}
= 2 + \frac{k + 1 - (2k + 4))}{2^{(k + 1)}}
= 2 - \frac{k + 3}{2^{(k + 1)}}        (terbukti)