Pertanyaan

Buktikan dengan induksi matematika. 2 1 + 2 2 2 + 2 3 3 + ⋯ + 2 n n = 2 − 2 n n + 2 berlaku untuk semua n bilangan asli.

Buktikan dengan induksi matematika.

$\frac{1}{2} + \frac{2}{2 ^{2}} + \frac{3}{2 ^{3}} + \dots + \frac{n}{2 ^{n}} = 2 - \frac{n + 2}{2 ^{n}}$

berlaku untuk semua n bilangan asli.

F. Kurnia

Master Teacher

Mahasiswa/Alumni Universitas Jember

Jawaban terverifikasi

Pembahasan

Misalkan . Langkah-langkah induksi matematika, yaitu sebagai berikut. Dibuktikan benar untuk Jadi, benar untuk diamsusikan benar untuk , sehingga benar untuk Karena ruas kiri = ruas kanan, maka benar untuk Dengan demikian, terbukti benaruntuk n bilangan asli.

Misalkan $begin mathsize 12px style P subscript n identical to 1 half plus 2 over 2 squared plus 3 over 2 cubed plus horizontal ellipsis plus n over 2 to the power of n equals 2 minus fraction numerator n plus 2 over denominator 2 to the power of n end fraction end style$ .

Langkah-langkah induksi matematika, yaitu sebagai berikut.

Dibuktikan P subscript n benar untuk n equals 1

$table attributes columnalign right center left columnspacing 0px end attributes row cell n over 2 to the power of n end cell equals cell 2 minus fraction numerator n plus 2 over denominator 2 to the power of n end fraction end cell row cell 1 over 2 to the power of 1 end cell equals cell 2 minus fraction numerator 1 plus 2 over denominator 2 to the power of 1 end fraction end cell row cell 1 half end cell equals cell 2 minus 3 over 2 end cell row cell 1 half end cell equals cell 1 half end cell end table$

Jadi, P subscript n benar untuk n equals 1

P subscript n diamsusikan benar untuk n equals k , sehingga

$begin mathsize 12px style 1 half plus 2 over 2 squared plus 3 over 2 cubed plus horizontal ellipsis plus k over 2 to the power of k equals 2 minus fraction numerator k plus 2 over denominator 2 to the power of k end fraction end style$

P subscript n benar untuk n equals k plus 1

$begin mathsize 12px style rightwards arrow Ruas space kiri equals 1 half plus 2 over 2 squared plus 3 over 2 cubed plus horizontal ellipsis plus k over 2 to the power of k plus fraction numerator k plus 1 over denominator 2 to the power of k plus 1 end exponent end fraction equals 2 minus fraction numerator k plus 2 over denominator 2 to the power of k end fraction plus fraction numerator k plus 1 over denominator 2 to the power of k plus 1 end exponent end fraction equals 2 plus fraction numerator negative 2 left parenthesis k plus 2 right parenthesis plus k plus 1 over denominator 2 to the power of k plus 1 end exponent end fraction equals 2 plus fraction numerator negative 2 k minus 4 plus k plus 1 over denominator 2 to the power of k plus 1 end exponent end fraction equals 2 plus fraction numerator negative k minus 3 over denominator 2 to the power of k plus 1 end exponent end fraction equals 2 plus fraction numerator negative left parenthesis k plus 3 right parenthesis over denominator k to the power of k plus 1 end exponent end fraction equals 2 minus fraction numerator k plus 3 over denominator k to the power of k plus 1 end exponent end fraction end style$

$begin mathsize 12px style rightwards arrow Ruas space kanan equals 2 minus fraction numerator left parenthesis k plus 1 right parenthesis plus 2 over denominator 2 to the power of left parenthesis k plus 1 right parenthesis end exponent end fraction equals 2 minus fraction numerator k plus 3 over denominator 2 to the power of k plus 1 end exponent end fraction end style$

Karena ruas kiri = ruas kanan, maka P subscript n benar untuk n equals k plus 1

Dengan demikian,

$begin mathsize 12px style 1 half plus 2 over 2 squared plus 3 over 2 cubed plus horizontal ellipsis plus n over 2 to the power of n equals 2 minus fraction numerator n plus 2 over denominator 2 to the power of n end fraction end style$

terbukti benar untuk n bilangan asli.