9 - Binomial Theorem Questions Answers

general term of the series = r(n+1-r)² = r(n+1)² + r³ - 2(n+1)r²

now put sigma then solve                                       

  ∑ r ⁿC_rx^ry^n-r

= ∑ r ⁿC_rx^r(1-x)^n-r

= ∑ r n!/r!(n-r)! x^r(1-x)^n-r

= n (n-1)!/(r-1)!(n-r)! x^r(1-x)^n-r 

= n ^(n-1)C_(r-1) x^r(1-x)^n-r

= n ^(n-1)C_r x^r+1(1-x)^n-r-1             let r = r+1

= nx  ^(n-1)C_r x^r(1-x)^n-r-1 

^{now solve} 

exact 0 and lim tends to 0 are different

There should be any variabe in the question 

(1 + 0.0001) ¹⁰⁰⁰⁰ = 1 + 10000*0.0001 + other terms = greater then 2

besides it lim _x->0 (1+x)^1/x = e = less then 3

it means the given expression is more then 2 but less then 3 

so 2 will be the answer

Please send the complete question

(1+λ)ⁿ = C₀ + C₁x + ..................................C_nxⁿ

similarly for (1+μ)ⁿ and 

(λ+μ)ⁿ = C₀λⁿ + C₁λ^n-1μ + ........................ + C_nμⁿ

On multiplying these 3, we will get the coefficient of  λⁿμⁿ = ∑C_n³      (Here you should remember that C₀ = C_n, C₁ = C_n-1 and etc)

For getting the value of  ∑C_n³ :

multiply the expansions of (1+x)ⁿ, (1-x)ⁿ and [1-(1/x²)]ⁿ and calculate the coefficient of x⁰, we get ∑C_n³

and if we multiply (1+x)ⁿ, (1-x)ⁿ and [1-(1/x²)]ⁿ, then we get (-1)ⁿ(1-x²)²ⁿ/x²ⁿ so coefficient of x⁰ in this expression will be (-1)^n 2nC_n (-1)ⁿ  = ²ⁿC_n

Let x = (5√5 + 11)²ⁿ⁺¹ and y = (5√5 - 11)²ⁿ⁺¹

so x*y = 4²ⁿ⁺¹ = even number

similarly x - y = integer

if we consider the value of y , it will be (5*2.3-5)²ⁿ⁺¹ = (0.5)²ⁿ⁺¹  = fraction number less than 1   (approx)

so definitely y will be the fractional part of x, and difference of x and y is even so integer part of x will be even

Total number of terms in the expansion = 4n-2+1 = 4n-1

for a negative real number i² is compulsory so associated terms are 3rd, 7th, 11th ..........

Total number of terms = n