हिंदी

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

Advertisements
Advertisements

प्रश्न

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

 
  
  
Advertisements

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.

 
 
shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 18: Binomial Theorem - Exercise 18.1 [पृष्ठ १२]

APPEARS IN

आरडी शर्मा Mathematics [English] Class 11
अध्याय 18 Binomial Theorem
Exercise 18.1 | Q 12 | पृष्ठ १२

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Expand the expression (1– 2x)5


Expand the expression: `(x/3 + 1/x)^5`


Expand the expression: `(x + 1/x)^6`


Using Binomial Theorem, evaluate the following:

(96)3


Using Binomial Theorem, evaluate of the following:
(102)5


Using binomial theorem, evaluate f the following:

(101)4


Using binomial theorem, evaluate the following:

(99)5


Prove that `sum_(r-0)^n 3^r  ""^nC_r = 4^n`


Find a if the coefficients of x2 and x3 in the expansion of (3 + ax)9 are equal.


Evaluate `(sqrt3 + sqrt2)^6 - (sqrt3 - sqrt2)^6`


Expand using Binomial Theorem `(1+ x/2 - 2/x)^4, x != 0`


Using binomial theorem determine which number is larger (1.2)4000 or 800?

 

Find the rth term in the expansion of `(x + 1/x)^(2r)`


Expand the following (1 – x + x2)4 


Evaluate: `(x^2 - sqrt(1 - x^2))^4 + (x^2 + sqrt(1 - x^2))^4`


Find the coefficient of x11 in the expansion of `(x^3 - 2/x^2)^12`


Determine whether the expansion of `(x^2 - 2/x)^18` will contain a term containing x10?


Find the term independent of x in the expansion of `(sqrt(x)/sqrt(3) + sqrt(3)/(2x^2))^10`.


If n is a positive integer, find the coefficient of x–1 in the expansion of `(1 + x)^2 (1 + 1/x)^n`


Find the coefficient of x50 after simplifying and collecting the like terms in the expansion of (1 + x)1000 + x(1 + x)999 + x2(1 + x)998 + ... + x1000 .


If (1 – x + x2)n = a0 + a1 x + a2 x2 + ... + a2n x2n , then a0 + a2 + a4 + ... + a2n equals ______.


Find the coefficient of x in the expansion of (1 – 3x + 7x2)(1 – x)16.


Find the coefficient of x15 in the expansion of (x – x2)10.


Find the sixth term of the expansion `(y^(1/2) + x^(1/3))^"n"`, if the binomial coefficient of the third term from the end is 45.


If the coefficient of second, third and fourth terms in the expansion of (1 + x)2n are in A.P. Show that 2n2 – 9n + 7 = 0.


Find the coefficient of x4 in the expansion of (1 + x + x2 + x3)11.


In the expansion of (x + a)n if the sum of odd terms is denoted by O and the sum of even term by E. Then prove that 4OE = (x + a)2n – (x – a)2n 


Given the integers r > 1, n > 2, and coefficients of (3r)th and (r + 2)nd terms in the binomial expansion of (1 + x)2n are equal, then ______.


The two successive terms in the expansion of (1 + x)24 whose coefficients are in the ratio 1:4 are ______.


If the coefficients of (2r + 4)th, (r – 2)th terms in the expansion of (1 + x)18 are equal, then r is ______.


Let `(5 + 2sqrt(6))^n` = p + f where n∈N and p∈N and 0 < f < 1 then the value of f2 – f + pf – p is ______. 


Share
Notifications

Englishहिंदीमराठी


      Forgot password?
Use app×