मराठी

Show that 2 4 N + 4 − 15 N − 16 , Where N ∈ N is Divisible by 225.

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प्रश्न

Show that  \[2^{4n + 4} - 15n - 16\]  , where n ∈  \[\mathbb{N}\]  is divisible by 225.

 
  
  
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उत्तर

We have,

\[2^{4n + 4} - 15n - 16 = 2^{4\left( n + 1 \right)} - 15n - 16\]

\[ = {16}^{n + 1} - 15n - 16\]

\[ = \left( 1 + 15 \right)^{n + 1} - 15n - 16\]

\[ =^{n + 1} C_0 {15}^0 +^{n + 1} C_1 {15}^1 +^{n + 1} C_2 {15}^2 + . . . +^{n + 1} C_{n + 1} {15}^{n + 1} - 15n - 16\]

\[ = 1 + (n + 1)15 +^{n + 1} C_2 {15}^2 + . . . +^{n + 1} C_{n + 1} {15}^{n + 1} - 15n - 16\]

\[ = 1 + 15n + 15 +^{n + 1} C_2 {15}^2 + . . . +^{n + 1} C_{n + 1} {15}^{n + 1} - 15n - 16\]

\[ =^{n + 1} C_2 {15}^2 + . . . +^{n + 1} C_{n + 1} {15}^{n + 1} \]

\[ = {15}^2 \left( {}^{n + 1} C_2 + . . . +^{n + 1} C_{n + 1} {15}^{n - 1} \right)\]

\[ = 225\left( {}^{n + 1} C_2 + . . . +^{n + 1} C_{n + 1} {15}^{n - 1} \right)\]

Thus, ​ 

\[2^{4n + 4} - 15n - 16\] , where n ∈  \[\mathbb{N}\]  is divisible by 225.

 
 
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पाठ 18: Binomial Theorem - Exercise 18.1 [पृष्ठ १२]

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आर.डी. शर्मा Mathematics [English] Class 11
पाठ 18 Binomial Theorem
Exercise 18.1 | Q 12 | पृष्ठ १२

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