Using binomial theorem, prove that 32n+2 – 8n – 9 is divisible by 64.​

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1. ⭐Question:–

   Using binomial theorem, prove that

   [tex] Using \: binomial \: theorem, \: prove \: that\: \\ 3 {}^{2n + 2 \: } – \: 8x \: – \: 9 \: is \: divisble \: by \: 64.[/tex]

   ⭐Answer:–

   Solution:-

   Let p (x) =

   [tex]3 {}^{2n + 2} \: – 8x \: – \: 9 \: is \: divisble \: by \: 64 \\ [/tex]

   [tex]when \: put \: n \: = \: 1 \\ [/tex]

   [tex]p \: (1) = \: 3 {}^{4} \: – 8 \: – 9 = \: 64[/tex]

   Let n = k and we get,

   [tex]p \: (k) \: = \: 3 {}^{2k + 2} \: – \: 8k \: – 9 \: divisble \: by \: 64[/tex]

   [tex]3 {}^{2k + 2} \: – 8k \: – 9 \: = \: 64m \: \: \: \: when \: m \: =N\\ [/tex]

   [tex]now \: we \: shall \: prove \: that \: p {}^{k + 1} is \: also \: true[/tex]

   [tex]p \: (k + 1) \: = \: 3 {}^{2(k + 1) + 2} \: – 8(k + 1) – 9 \: is \\ divisble \: by \: 64[/tex]

   Now,

   [tex]p(k + 1) = \: 3 {}^{2(k + 1) + 2} – 8(k + 1) \: – 9 = 3 {}^{2} .3 {}^{2k + 2} – 8k \: – 17[/tex]

   [tex] = \: 9.3 {}^{2k + 2} – 8k – 17[/tex]

   [tex] = \: 9(64m + 8k + 9) \: – 8k \: – 17[/tex]

   [tex] = 9.64m \: 72k + 81 – 8k – 17[/tex]

   [tex] = 9.64m + 64k + 64[/tex]

   [tex] = 64(9m + k + 1)[/tex] which is disability by 64.

   Thus, p (k+1) is true whenever p(k) is true.

   Hence, by principle mathematical induction,

   px is true for all natural number.

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