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Prove that 1 3 n + 6 n − 1 is divisible by 7 .

Prove that  is divisible by .

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N. Puspita

Master Teacher

Jawaban terverifikasi

Jawaban

terbukti bahwa habis dibagi 7.

terbukti bahwa P subscript n identical to 13 to the power of n plus 6 to the power of n minus 1 end exponent habis dibagi 7.

Pembahasan

Pembahasan
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Misalkan habis dibagi Langkah pembuktian Langkah pertama: habis dibagi Langkah kedua: Anggap habis dibagi . Akan dibuktikan kebenaran dari pernyataan habis dibagi . perhatikan habis dibagi 7, dan juga habis dibagi 7. Maka, habis dibagi 7. Jadi, terbukti bahwa habis dibagi 7.

Misalkan P subscript n identical to 13 to the power of n plus 6 to the power of n minus 1 end exponent habis dibagi 7

Langkah pembuktian

Langkah pertama:

P subscript 1 equals 13 to the power of 1 plus 6 to the power of 1 minus 1 end exponent equals 13 plus 1 equals 14 equals 2 space. space 7  habis dibagi 7

Langkah kedua:

Anggap P subscript k identical to 13 to the power of k plus 6 to the power of k minus 1 end exponent habis dibagi 7. Akan dibuktikan kebenaran dari pernyataan 

P subscript k plus 1 end subscript identical to 13 to the power of k plus 1 end exponent plus 6 to the power of open parentheses k plus 1 close parentheses minus 1 end exponent habis dibagi 7

13 to the power of k plus 1 end exponent plus 6 to the power of open parentheses k plus 1 close parentheses minus 1 end exponent equals 13 to the power of k plus 1 end exponent plus 6 to the power of k space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 13 to the power of k. space 13 plus 6 to the power of k space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 6 space. space 13 to the power of k plus 7 space. space 13 to the power of k plus 6 to the power of 6 k end exponent space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 6 open parentheses 13 to the power of k plus 6 to the power of k over 6 close parentheses plus 7 space. space 13 to the power of k space space space space space space space space space space space space space space space space space space space space space space space space space space space space equals 6 open parentheses 13 to the power of k plus 6 to the power of k minus 1 end exponent close parentheses plus 7 space. space 13 to the power of k 

perhatikan 13 to the power of k plus 6 to the power of k minus 1 end exponent habis dibagi 7, dan 7 space. space 13 to the power of k juga habis dibagi 7. Maka, P subscript k plus 1 end subscript identical to 13 to the power of k plus 1 end exponent plus 6 to the power of open parentheses k plus 1 close parentheses minus 1 end exponent habis dibagi 7. 

Jadi, terbukti bahwa P subscript n identical to 13 to the power of n plus 6 to the power of n minus 1 end exponent habis dibagi 7.

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