Advertisements
Advertisements
प्रश्न
If n is an odd positive integer, prove that the coefficients of the middle terms in the expansion of (x + y)n are equal
Advertisements
उत्तर
Given n is odd.
So let n = 2n + 1
Where n is an integer.
The expansion (x + y)n has n + 1 terms.
= 2n + 1 + 1
= 2(n + 1) terms which is an even number.
So the middle term are `("t"^2("n"+ 1))/2` = tn+1
`"t"_(2("n" + 1))` = tn+1 and `"t"_("n" + + 1)` = tn+2
(i.e.) The middle terms are tn+1 and tn+2
tn+1 = `""^(2"n" + 1)"C"_"n"` and tn+2
= `"t"_(n" + 1 + 1)`
= 2n + 1Cn+1
Now n + n + 1 = 2n + 1
⇒ `""^(2"n" + 1)"C"_"n" = ""^(2"n" + 1)"C"_("n" + 1)`
⇒ The coefficient of the middle terms in (x + y)n are equal.
APPEARS IN
संबंधित प्रश्न
Evaluate the following using binomial theorem:
(999)5
Expand the following by using binomial theorem.
`(x + 1/x^2)^6`
Find the middle terms in the expansion of
`(x + 1/x)^11`
Find the middle terms in the expansion of
`(2x^2 - 3/x^3)^10`
Find the term independent of x in the expansion of
`(x^2 - 2/(3x))^9`
Show that the middle term in the expansion of is (1 + x)2n is `(1*3*5...(2n - 1)2^nx^n)/(n!)`
Find the Co-efficient of x11 in the expansion of `(x + 2/x^2)^17`
The constant term in the expansion of `(x + 2/x)^6` 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
Expand `(2x^2 - 3/x)^3`
Using binomial theorem, indicate which of the following two number is larger: `(1.01)^(1000000)`, 10
If n is a positive integer, using Binomial theorem, show that, 9n+1 − 8n − 9 is always divisible by 64
If n is a positive integer and r is a non-negative integer, prove that the coefficients of xr and xn−r in the expansion of (1 + x)n are equal
If a and b are distinct integers, prove that a − b is a factor of an − bn, whenever n is a positive integer. [Hint: write an = (a − b + b)n and expaand]
Choose the correct alternative:
The remainder when 3815 is divided by 13 is
