Pertanyaan

Prove that the following are true for all positive integral value of n. Use mathematical induction. 2. a + a ⋅ r + a ⋅ r 2 + ... + a ⋅ r n − 1 = 1 − r a ( 1 − r n )

Prove that the following are true for all positive integral value of n. Use mathematical induction.

2. $a + a \cdot r + a \cdot r^{2} + ... + a \cdot r^{n - 1} = \frac{a ( 1 - r ^{n} )}{1 - r}$

A. Acfreelance

Master Teacher

Mahasiswa/Alumni UIN Walisongo Semarang

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terbukti bahwa karena hasil sisi kiri dan kanan sama

terbukti bahwa $straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of straight n minus 1 end exponent equals fraction numerator straight a open parentheses 1 minus straight r to the power of straight n close parentheses over denominator 1 minus straight r end fraction space$ karena hasil sisi kiri dan kanan sama

Pembahasan

Membuktikan dengan cara induksi matematika dimana untuk nilai n =1 maka Untuk n = k diasumsikan terbukti maka Untuk n = k+1 maka Jadi terbukti bahwa karena hasil sisi kiri dan kanan sama

Membuktikan dengan cara induksi matematika dimana untuk nilai n =1 maka

Untuk n = k diasumsikan terbukti maka

Untuk n = k+1 maka

$table attributes columnalign right center left columnspacing 0px end attributes row cell straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of straight n minus 1 end exponent end cell equals cell fraction numerator straight a open parentheses 1 minus straight r to the power of straight n close parentheses over denominator 1 minus straight r end fraction space end cell row cell straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of straight k minus 1 end exponent plus straight a. straight r to the power of straight k plus 1 minus 1 end exponent end cell equals cell fraction numerator straight a open parentheses 1 minus straight r to the power of straight k plus 1 end exponent close parentheses over denominator 1 minus straight r end fraction end cell row cell fraction numerator straight a open parentheses 1 minus straight r to the power of straight k close parentheses over denominator 1 minus straight r end fraction plus ar to the power of straight k end cell equals cell space fraction numerator straight a open parentheses 1 minus straight r to the power of straight k plus 1 end exponent close parentheses over denominator 1 minus straight r end fraction end cell row cell fraction numerator straight a minus ar to the power of straight k plus left parenthesis 1 minus straight r right parenthesis ar to the power of straight k over denominator 1 minus straight r end fraction end cell equals cell space fraction numerator straight a open parentheses 1 minus straight r to the power of straight k plus 1 end exponent close parentheses over denominator 1 minus straight r end fraction end cell row cell fraction numerator straight a open parentheses 1 minus straight r to the power of straight k plus 1 end exponent close parentheses over denominator 1 minus straight r end fraction end cell equals cell space fraction numerator straight a open parentheses 1 minus straight r to the power of straight k plus 1 end exponent close parentheses over denominator 1 minus straight r end fraction rightwards arrow terbukti end cell end table$

Jadi terbukti bahwa $straight a plus straight a times straight r plus straight a times straight r squared plus... plus straight a times straight r to the power of straight n minus 1 end exponent equals fraction numerator straight a open parentheses 1 minus straight r to the power of straight n close parentheses over denominator 1 minus straight r end fraction space$ karena hasil sisi kiri dan kanan sama

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