Let α=∑_(r=0)^n▒〖(4r^2+2r+1)n_(C_r ) 〗 and β=(∑_(r=0)^n▒n_(C_r )/(r+1))+1/(n+1).

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Let alpha= Sigma r = 0 ^ n (4r^ 2 +2r+1)^ n C r and 140 (2alpha)/beta 281 , || Let Let alpha= Sigma r = 0 ^ n (4r^ 2 +2r+1)^ n C r and 140 (2alpha)/beta 281 , || Let's IQRA JEE
Let a=sigma (4r^2+2r+1)nCr and beta=sigma nCr/(r+1)+1/n+1 if 2alpha/beta €(140,281),then value n is Let a=sigma (4r^2+2r+1)nCr and beta=sigma nCr/(r+1)+1/n+1 if 2alpha/beta €(140,281),then value n is
Let α = ∑ (4r² +2r +1) ⁿCᵣ & β = (∑ ⁿCᵣ/r+1) + 1/n+1. if 140 ≤ 2α/β ≤ 281 then the value of n is Let α = ∑ (4r² +2r +1) ⁿCᵣ & β = (∑ ⁿCᵣ/r+1) + 1/n+1. if 140 ≤ 2α/β ≤ 281 then the value of n is
If (4x^(2) + 1)^(n) = sum_(r=0)^(n)a_(r)(1+x^(2))^(n-r)x^(2r), then the value of sum_(r=0)^(n)a_... If (4x^(2) + 1)^(n) = sum_(r=0)^(n)a_(r)(1+x^(2))^(n-r)x^(2r), then the value of sum_(r=0)^(n)a_...
et α=∑nr=0(4r2+2r+1)nCr and β=(∑nr=0nCrr+1)+1n+1 . If 140 2α/β 281 , then the value of n i et α=∑nr=0(4r2+2r+1)nCr and β=(∑nr=0nCrr+1)+1n+1 . If 140 2α/β 281 , then the value of n i
If sigma r=0 to 5 11C(2r+1)/2r+2 = m/n, gcd(n,n)=1, then n-n is equal to If sigma r=0 to 5 11C(2r+1)/2r+2 = m/n, gcd(n,n)=1, then n-n is equal to
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