Advertisements
Advertisements
प्रश्न
If n is a positive integer, using Binomial theorem, show that, 9n+1 − 8n − 9 is always divisible by 64
Advertisements
उत्तर
(1 + x)n = nC0 + nC1x + nC2x2 + ........ + nCn−1 xn−1 + nCnxn
Put x = 8 we get
(1 + 8)n = nC0 + nC1(8) + nC2(8)2 + ......... + nCn–1 8n–1 + nCn . 8n
9n = nC0 + nC1(8) + nC2(8)2 + ......... + nCn–1 8n–1 + nCn . 9n = 1 + 8n + nC2 × 82 + ........ +
nCn–1 8n–1 + nCn . 8n
9n - 8n – 1 = nC2 × 82 + ........ + nCn–1 8n–1 + nCn × 8n
9n - 8n – 1 = 82 [nC2 + .......... + nCn–1 × 8n–3 + nCn × 8n–2]
Which is divisible by 64 for all positive integer n.
∴ 9n – 8n – 1 is divisible by 64 for all positive integer n.
Put n = n + 1 we get
9n + 1 – 8 (n + 1) – 1 is divisible by 64 for all possible integer n
(9n + 1 – 8n – 8 – 1) is divisible by 64
∴ 9n + 1 – 8n – 9 is always divisible by 64
APPEARS IN
संबंधित प्रश्न
Evaluate the following using binomial theorem:
(999)5
Find the 5th term in the expansion of (x – 2y)13.
Find the middle terms in the expansion of
`(x + 1/x)^11`
Find the term independent of x in the expansion of
`(x^2 - 2/(3x))^9`
Find the term independent of x in the expansion of
`(x - 2/x^2)^15`
Find the term independent of x in the expansion of
`(2x^2 + 1/x)^12`
The constant term in the expansion of `(x + 2/x)^6` is
The last term in the expansion of (3 + √2 )8 is:
Sum of binomial coefficient in a particular expansion is 256, then number of terms in the expansion is:
Sum of the binomial coefficients is
Find the coefficient of x2 and the coefficient of x6 in `(x^2 -1/x^3)^6`
Find the coefficient of x4 in the expansion `(1 + x^3)^50 (x^2 + 1/x)^5`
In the binomial expansion of (1 + x)n, the coefficients of the 5th, 6th and 7th terms are in AP. Find all values of n
Prove that `"C"_0^2 + "C"_1^2 + "C"_2^2 + ... + "C"_"n"^2 = (2"n"!)/("n"!)^2`
Choose the correct alternative:
The value of 2 + 4 + 6 + … + 2n is
Choose the correct alternative:
The remainder when 3815 is divided by 13 is
