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BINOMIAL THEOREM — MATHS STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceMATHSBINOMIAL THEOREM3 Marks
Question
Show that $9^{n+1} - 8n - 9$ is divisible by $64$ whenever $n$ is a positive integer.
✓

Answer

We have $9^{n+1} = (1 + 8)^{n+1}$
$ ={ }^{n+1} C_0+{ }^{n+1} C_1(8)+{ }^{n+1} C_2(8)^2+{ }^{n+1} C_3(8)^3+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1}$
$=1+(n+1) \times 8+{ }^{n+1} C_2(8)^2+{ }^{n+1} C_3(8)^3+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1}$
$=1+8 n+8+{ }^{n+1} C_2(8)^2+{ }^{n+1} C_3(8)^3+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1}$
$=9+8 n+{ }^{n+1} C_2(8)^2+{ }^{n+1} C_3(8)^3+\ldots \ldots \ldots+{ }^{n+1} C_{n+1}(8)^{n+1}$
$\Rightarrow 9^{n+1}-8 n-9={ }^{n+1} C_2(8)^2+{ }^{n+1} C_3(8)^3+\ldots+{ }^{n+1} C_{n+1}(8)^{n+1}$
$=64\left[{ }^{n+1} C_2+{ }^{n+1} C_3 \cdot 8+\ldots+{ }^{n+1} C_{n+1} \cdot 8^{n+1}\right] $
which show that $9^{n+1}-8^{n-9}$ is divisible by $64$ wherever $n$ is a positive integer

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