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Question
Using binomial theorem, prove that \[3^{2n + 2} - 8n - 9\] is divisible by 64, \[n \in N\] .
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Solution
\[3^{2n + 2} - 8n - 9 = 9^{n + 1} - 8n - 9 . . . \left( 1 \right)\]
Consider
\[9^{n + 1} = \left( 1 + 8 \right)^{n + 1} \]
\[ \Rightarrow 9^{n + 1} =^{n + 1}{}{C}_0 \times 8^0 + ^{n + 1}{}{C}_1 \times 8^1 +^{n + 1}{}{C}_2 \times 8^2 + ^{n + 1}{}{C}_3 \times 8^3 + . . . +^{n + 1}{}{C}_{n + 1} \times 8^{n + 1}\]
\[\Rightarrow 9^{n + 1} = 1 + 8(n + 1) + [^{n + 1}{}{C}_2 \times 8^2 + ^{n + 1}{}{C}_3 \times 8^3 + . . . +^{n + 1}{}{C}_{n + 1} \times 8^{n + 1} ]\]
\[ \Rightarrow 9^{n + 1} - 8n - 9 = 64( ^{n + 1}{}{C}_2 + ^{n + 1}{}{C}_3 \times 8^1 + . . . + ^{n + 1}{}{C}_{n + 1} \times 8^{n - 1} ]\]
\[ \Rightarrow 9^{n + 1} - 8n - 9 = 64 \times\text{ An integer } \]
\[ 9^{n + 1} - 8n - 9 \text{ is divisible by } 64\]
\[\text{ Or, } \]
\[ 3^{2n + 2} - 8n - 9 \text{ is divisible by } 64 \left[ \text{ From } (1) \right]\]
\[\text{ Hence proved .}\]
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