हिंदी

Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer. - Mathematics

Advertisements
Advertisements

प्रश्न

Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer.

योग
Advertisements

उत्तर

We have to prove that 9n + 1 - 8n - 9 = 64k

∴ 9 n+1 - 8n - 9 = (8 + 1)n+ 1 - 8n - 9     [put 9 = 8 + 1]

= \[\ce{[^{n + 1}C_0 8^{n + 1} + ... + ^{n + 1}C_{n-2} 8^3 + ^{n + 1} C_{n - 1} 8^2 + ^{n + 1} C_n 8 + ^{n + 1}C_{n + 1}] - 8n - 9}\]

= \[\ce{^{n + 1} C_0 8 ^{n + 1} + ... + ^{n + 1}C_{n - 2} 8^3 + ^{n + 1}C_{n - 1}8^2 + (n + 1) 8 + 1 - 8n - 9}\]

= \[\ce{^{n + 1}C_0 8 ^{n + 1} + ... + ^{n + 1}C_{n - 2}8^3  + ^{n + 1}C_{n - 1}8^2 + 8n + 8 + 1 - 8n - 9}\]

= \[\ce{^{n + 1}C_0 8^{n + 1} + ... + ^{n + 1}C_{n - 2}8^3 + ^{n + 1}C_{n - 1}8^2}\]

= \[\ce{8^2 [^{n + 1}C_0 8^{n - 1} + ... + ^{n + 1}C_{n - 2} 8 + ^{n + 1}C_{n - 1}]}\]

= 64k [where, k = n + 1C0 8n - 1 + .... + n + 1Cn - 1]

Hence, 9n + 1 - 8n - 9 is divisible by 64, whenever n is a positive integer.

shaalaa.com
  क्या इस प्रश्न या उत्तर में कोई त्रुटि है?
अध्याय 8: Binomial Theorem - Exercise 8.1 [पृष्ठ १६७]

APPEARS IN

एनसीईआरटी Mathematics [English] Class 11
अध्याय 8 Binomial Theorem
Exercise 8.1 | Q 13 | पृष्ठ १६७

वीडियो ट्यूटोरियलVIEW ALL [1]

संबंधित प्रश्न

Expand the expression: (1– 2x)5


Expand the expression: `(2/x - x/2)^5`


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


Using binomial theorem, evaluate f the following:

(101)4


Using binomial theorem, evaluate the following:

(99)5


Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000.


Find (a + b)4 – (a – b)4. Hence, evaluate `(sqrt3 + sqrt2)^4 - (sqrt3 - sqrt2)^4`


Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate `(sqrt2 + 1)^6 + (sqrt2 -1)^6`


Find ab and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively.


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


Find the value of `(a^2 + sqrt(a^2 - 1))^4 + (a^2 - sqrt(a^2 -1))^4`


Find an approximation of (0.99)5 using the first three terms of its expansion.


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


If n is a positive integer, prove that \[3^{3n} - 26n - 1\]  is divisible by 676.

 
 

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

 

Find the value of (1.01)10 + (1 − 0.01)10 correct to 7 places of decimal.

 

Expand the following (1 – x + x2)4 


Find the 4th term from the end in the expansion of `(x^3/2 - 2/x^2)^9`


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`


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 a1, a2, a3 and a4 are the coefficient of any four consecutive terms in the expansion of (1 + x)n, prove that `(a_1)/(a_1 + a_2) + (a_3)/(a_3 + a_4) = (2a_2)/(a_2 + a_3)`


The coefficient of xp and xq (p and q are positive integers) in the expansion of (1 + x)p + q are ______.


If z = `sqrt(3)/2 + i^5/2 + sqrt(3)/2 - i^5/2`, then ______.


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.


The total number of terms in the expansion of (x + a)100 + (x – a)100 after simplification is ______.


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 ______.


The sum of the last eight coefficients in the expansion of (1 + x)16 is equal to ______.


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×