Test: Binomial Theorem- 2 - JEE MCQ

/* We know that,
If a,b,c are in A.P then
a+c = 2b */

/* Splitting the middle term, we get

Here , n=2 is not possible because number of terms are more than three (According to the problem )
Therefore, n = 7

T_(r+1) = ¹¹C_r x^(11-2r) (m/r)^r (-1)^r
= ¹¹C_r x^(11-2r) m^r (-1)^r
Coefficient of x³ = 11 - 2r = 3
8 = 2r 
r = 4
T5 = ¹¹C₄ x³ m⁴ (-1)
Coefficient of x³ = ¹¹C₄ m⁴
= 330 m⁴

The coefficient of x¹⁷ is given by 
−1 + (−2) + (−3) + ….. (−18)
= −1 − 2 − 3….. − 18
= − (18(18+1))/2
= − 9(19)
= − 171

T_r+1 = ²ⁿCr(x)^r (1)^2n-r
T_r+3 = ²ⁿC_r+2 (x)^r+2 (1)^2n-r-2
T_r+1 : T_r+3 
= ²ⁿC_r = ²ⁿC_r+2
=> 2n!/(2n-r)!r! = 2n!/(r+2)!(2n-r-2)!
=> 1/(2n-r)(2n-r-1) = 1/(r+2)(r+1)
=> (r+2)(r+1) = (2n-r)(2n-r-1)
=> r2 + 3r + 2 = 4n2 - 2nr - 2n - 2nr + r^2 + r
=> 0 = 4n2 - 4nr - 2n + r - 3r - 2
=> 0 = 4n2 - 4nr - 2n + r - 3r - 2 - 2 + 2
=> 0 = 4n2 - 4nr - 4 - [2n + 2r - 2]
=> 4n(n-r-1) -2(n-r-1)
Therefore n - r - 1 = 0
=> n = r+1

 (1+x)ⁿ= nC0+ nC1x+ nC2x²+...+ nC0xⁿ ....(1)
(1−x)ⁿ= nC0 − nC1x+ nC2x²−...+ nC0 xⁿ....(2)
⇒(1+x)ⁿ−(1−x)ⁿ = nC0+ nC1x+ nC2x²+...+ nC0xⁿ− nC0+ nC1x− nC2x²+...+ nC0xⁿ
 =2(nC1x+nC3x³+....+xⁿ)

Given: n=60 and put x=1
⇒(1+1)⁶⁰−(1−1)⁶⁰
 =2⁶⁰
⇒2⁶⁰=2(nC1x+nC3x³+....+xⁿ)

∴Sum of odd powers of x in the expansion is = nC1x+nC3x³+....+xⁿ =2⁶⁰−1 =2⁵⁹

In case of negative or fractional power, expansion (1+x)^n is valid only when |x| < 1
(6 - 3x)^-1/2
= (6^-1/2 (1 - x/2)^-1/2)
So, this equation exists only when |x/2| < 1
|x| < 2

Middle term in the expansion of (1+x)²ⁿ; is 
=tn+1 = 2nCn.1⁽²ⁿ⁻ⁿ⁾.xⁿ
= {(2n)!.xⁿ}/(2n−n)!n!
= {{2n(2n−1)(2n−2)(2n−3)....4×3×2×1}/n! n!}/xⁿ
= {{2ⁿ[n(n−1)(n−2)....×2×1][(2n−1)(2n−3)....3×1]}/n! n!}xⁿ
= [[(2n−1)(2n−3)....3×1]/n!] 2ⁿxⁿ

(x+1)(x+3)(x+5)(x+7)...(x+199)
We have (x+1)(x+3)=x²+(1+3)x+3
(x+1)(x+3)(x+5)=x^3+(1+3+5)x² +(3+15+5)x+15
here Maximum power of x can be 100
For x^99 we have to multiply x from 99 terms and constant from remaning 1 term
So, when constant will be taken from (x+1) coff. of $$x⁹⁹ will be 1
when constant term will be taken from (x+3) coff. of x⁹⁹ will be 3
Similarly, When constant term will be taken from (x+199) coff. of x⁹⁹ will be 199.
∴ coefficient of x^99will be sum of all these coffecient=1+3+5+..+199

T_(r+1) =  ⁿC_r x^n-r a^r
T_(r+1) =  ^n-5Cr xn-5-r (1/x⁴)^r
= ^n-5C_r x^n-5-r (1/x^4r)
= ^n-5C_r x^n-5-5r 
=> n - 5 - 5r = 2r
=> n - 2r = 5(r + 1)