JEE Advanced Level Test: Binomial Numbers- 1 - JEE MCQ

Sum of odd terms in the expansion of (1+x)ⁿ will be 2^(n-1) ....(i)
Now 
(1−x²)ⁿ = ⁿC₀ − ⁿC₁ x2 + ⁿC₂ x4 + ....(−1)^{n n}C_n x²ⁿ .....(n)
(1+x²)ⁿ = ⁿC⁰ + _nC₁ x² + ⁿC₂ x⁴ +... nCn x²ⁿ .....(m)
Subtracting n from m, we get 
(1+x²)ⁿ − (1−x²)ⁿ = 2[ⁿC₁ x² + ⁿC₃ x⁶ + .....]
Now let x=1.
Hence we get 
2ⁿ − 0 = 2[ⁿC₁ + ⁿC₃ + .....]
Or ⁿC₁ + nC3x² + ⁿC₅ + ... = 2^(n-1)
Hence a = b
Now (1−x2)ⁿ
=(1+x)ⁿ (1−x)ⁿ
→[(a+b)(a−b)]
=a² − b².

Since T_{r + 1} = ⁿC_r a^{n – r} x^r in expansion of (a + x)ⁿ,
Therefore,

1/1!(n−1)! + 1/3!(n−3) + 1/5!(n−5) +,
1/n![nC1 + nC3 + nC5 + …………….]
= 1/n![2^(n-1)]

(2x + 1/x)ⁿ
Sum of binomial coefficient = 2ⁿ = 256
2ⁿ = 2⁸   => n=8
8C4(2x)⁴(1/x)⁴
= [(8*7*6*5) * (2⁴* x⁴-4)]/1*2*3*4
= 1120

p→ integer   f→ fraction 
(5+2(6)^½)ⁿ = p+f  {0<f<1}
f2 – f + pf – p 
f(f-1) +p(f-1)
(f+p) (f-1)
(p+f) = (5+2(6)^½)ⁿ = nC0 5ⁿ + nC1 5^(n-1)(2(6)^½) + nC2 5^(n-2)(2(6)^½) + nC3 5^(n-3)(2(6)^½) + ………………….nCn 5⁰ (2(6)^½)  --------------------------(1)
0 < (5+2(6)^½)ⁿ which is approx 
(5- 4.8989) = (0.11) < 1
0 <(0.11)ⁿ <1
0 <(5-2(6)^½)ⁿ <1
0 <f’ <1
f’ = (5+2(6)^½)ⁿ = nC0 5ⁿ - nC1 5^(n-1)(2(6)^½) + nC2 5^(n-2)(2(6)^½) - nC3 5^(n-3)(2(6)^½) + ………………….nCn 5⁰ (2(6)^½)  --------------------------(2)
Adding (1) & (2)
p+f+f’ = 2[= (nC0 5ⁿ + nC2 5^(n-2)(2(6)^½) + nC4 5^(n-4)(2(6)^½) + …………………]
p+f=f’ = 2K    K is integer
f+f’ = 2K-p = integer
f+f’ should be an integer
f² - f + pf - p 
= f(f-1) +p(f-1)
= (f+p) (f-1)   = -(p+f) (1-f)
= -(p+f) (f’)
= - (5+2(6)^½)ⁿ (5-2(6)^½)ⁿ
= - {(5+2(6)^½)  (5-2(6)^½)}ⁿ
= - (5)² - (2(6)²)²
= - [25 - 24]ⁿ  => - (1)ⁿ
= - 1 (which is negative integer)