Let α = ∑ (4r² +2r +1) ⁿCᵣ & β = (∑ ⁿCᵣ/r+1) + 1/n+1. if 140 ≤ 2α/β ≤ 281 then the value of n is

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Download lagu Let α = ∑ (4r² +2r +1) ⁿCᵣ & β = (∑ ⁿCᵣ/r+1) + 1/n+1. if 140 ≤ 2α/β ≤ 281 then the value of n is secara gratis hanya untuk keperluan promosi. Dukung artis favorit kamu dengan membeli musik original di iTunes atau platform resmi lainnya.

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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
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
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). 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).
Σ m from 1 to ∞ of Binomial(3m,m)/8ᵐ & Σ Σ m,n from 1 to ∞ of (αᵐ—1)/((m + 1) nᵐ⁺¹ βᵐ) Σ m from 1 to ∞ of Binomial(3m,m)/8ᵐ & Σ Σ m,n from 1 to ∞ of (αᵐ—1)/((m + 1) nᵐ⁺¹ βᵐ)
Let ` sum_(r=1)^n = f(n)`. Then `sum_(r =1)^n (2r-1)^4` is equal to : Let ` sum_(r=1)^n = f(n)`. Then `sum_(r =1)^n (2r-1)^4` is equal to :
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Let α=∑r=0n4r2+2r+1Crn and β=∑r=0nCrnr+1+1n+1. If 140<2αβ&.... Let α=∑r=0n4r2+2r+1Crn and β=∑r=0nCrnr+1+1n+1. If 140<2αβ&....