Let α=∑_ r=0 ^n(4 r^2+2 r+1) ^n C_r and β=(∑_ r=0 ^n ^n C_r r+1 )+(1)/(n+1) . If 140 < (2 α)/(β) < 281 , then the value of n is _______

Correct answer: 0 aligned & α=∑_ r=0 ^n(4 r^2+2 r+1) · ^n C_r & α=4 ∑_ r=0 ^n r^2 · (n)/(r) · ^ n-1 C_ r-1 +2 ∑_ r=0 ^n r · (n)/(r) · ^ n-1 C_ r- +∑_ r=0 ^n ^n C_r & +4 n ∑_ r=0 ^n ^ n-1 C_ r-1 +2 n ∑_ r=0 ^n ^ n-1 C_ r-1 +∑_ r=0 ^n ^n C_r & α=4 n(n-1) · 2^ n-2 +4 n · 2^ n-1 +2 n · 2^ n-1 +2^n & α=2^ n-2 [4 n(n-1)+8 n+4 n+4] & α=2^ n-2 [4 n^2+8 n+4] & α=2 n(n+1)^2 & β=∑_ r=0 ^n ^n C_r r+1 +(1)/(n+1) & =∑_ r=0 ^n C^ n+1 C_ r+1 n+1 +(1)/(n+1) & =(1)/(n+1)(1+ ^ n+1 C_1+ .+ ^ n+1 C_ n+1 ) & = 2^ n+1 n+1 & (2 α)/(β)= 2^ n+1 (n+1)^2 2^ n+1 ·(n+1)=(n+1)^3 & 140 < (n+1)^3 < 281 & n=4 (n+1)^3=125 & n=5 (n+1)^3=216 & n=6 (n+1)^3=343 & n=5 aligned

JEE Main 20248 AprilMorning Shift

Exam: JEE Main
Year: 2024
Date: 8 April
Shift: Morning Shift
Subject: Mathematics
Chapter: Binomial Theorem

JEE Main 2024 · 8 April · Morning Shift · Mathematics · Binomial Theorem

Let α=∑_ r=0 ^n(4 r^2+2 r+1) ^n C_r and β=(∑_ r=0 ^n ^n C_r r+1...

Let

\alpha=\sum_{r=0}^n\left(4 r^2+2 r+1\right){ }^n C_r

and

\beta=\left(\sum_{r=0}^n \frac{{ }^n C_r}{r+1}\right)+\frac{1}{n+1}

. If

140 < \frac{2 \alpha}{\beta} < 281

, then the value of

n

is _______

Correct answer

Step-by-step solution

\begin{aligned} & \alpha=\sum_{r=0}^n\left(4 r^2+2 r+1\right) \cdot{ }^n C_r \\ & \alpha=4 \sum_{r=0}^n r^2 \cdot \frac{n}{r} \cdot{ }^{n-1} C_{r-1}+2 \sum_{r=0}^n r \cdot \frac{n}{r} \cdot{ }^{n-1} C_{r-}+\sum_{r=0}^n{ }^n C_r \\ & +4 n \sum_{r=0}^n{ }^{n-1} C_{r-1}+2 n \sum_{r=0}^n{ }^{n-1} C_{r-1}+\sum_{r=0}^n{ }^n C_r \\ & \alpha=4 n(n-1) \cdot 2^{n-2}+4 n \cdot 2^{n-1}+2 n \cdot 2^{n-1}+2^n \\ & \alpha=2^{n-2}[4 n(n-1)+8 n+4 n+4] \\ & \alpha=2^{n-2}\left[4 n^2+8 n+4\right] \\ & \alpha=2 n(n+1)^2 \\ & \beta=\sum_{r=0}^n \frac{{ }^n C_r}{r+1}+\frac{1}{n+1} \\ & =\sum_{r=0}^n \frac{C^{n+1} C_{r+1}}{n+1}+\frac{1}{n+1} \\ & =\frac{1}{n+1}\left(1+{ }^{n+1} C_1+\ldots .+{ }^{n+1} C_{n+1}\right) \\ & =\frac{2^{n+1}}{n+1} \\ & \frac{2 \alpha}{\beta}=\frac{2^{n+1}(n+1)^2}{2^{n+1}} \cdot(n+1)=(n+1)^3 \\ & 140 < (n+1)^3 < 281 \\ & n=4 \Rightarrow(n+1)^3=125 \\ & n=5 \Rightarrow(n+1)^3=216 \\ & n=6 \Rightarrow(n+1)^3=343 \\ & \therefore n=5\end{aligned}

Practice Binomial Theorem on Quantrex Academy →

More from Binomial Theorem

If 26((2^3)/(3) 12 2 + (2^5)/(5) 12 4 + (2^7)/(7) 12 6 + … + 2^ 13 13 12 12 ) = 3^ 13 - α… JEE Main 2026 If (1 - x^3)^ 10 = ∑ _ r=0 ^ 10 a_r x^r (1-x)^ 30-2r , then 9a_9 a_ 10 is equal to… JEE Main 2026 If the coefficients of the middle terms in the binomial expansions of (1 + α x)^ 26 and… JEE Main 2026 The coefficient of x^2 in the expansion of (2x^2 + (1)/(x))^ 10 , x ≠ 0 , is : JEE Main 2026 If the sum of the coefficients of x^7 and x^ 14 in the expansion of ((1)/(x^3) - x^4)^n… JEE Main 2026 In the expansion of (9x- 1 3 x )^ 18 , x>0 , if the term independent of x is (221)k… JEE Main 2026 Let the smallest value of k ∈ ℕ , for which the coefficient of x^3 in (1+x)^3 + (1+x)^4 +… JEE Main 2026 If for 3 ≤ r ≤ 30 , 30 30-r + 3 30 31-r + 3 30 32-r + 30 33-r = m r , then m equals: JEE Main 2026 The number of elements in the set S = \ (r, k) : k ∈ ℤ and ^ 36 C_ r+1 = 6( ^ 35 C_r)… JEE Main 2026 Given below are two statements : Statement I: 25^ 13 +20^ 13 +8^ 13 +3^ 13 is divisible… JEE Main 2026 The sum of the coefficients of x^ 499 and x^ 500 in (1+x)^ 1000 +x(1+x)^ 999 +x^ 2 (1+x)^… JEE Main 2026 Let S=(1)/(25!)+(1)/(3!23!)+(1)/(5!21!)+… up to 13 terms. If 13 ~S= 2^ k n! , k ∈ ~N… JEE Main 2026 The value of ^ 100 C_ 50 51 + ^ 100 C_ 51 52 +….+ ^ 100 C_ 100 101 is : JEE Main 2026 The sum of all possible values of n ∈ N , so that the coefficients of x, x^ 2 and x^ 3 in… JEE Main 2026 Let C_ r denote the coefficient of x^ r in the binomial expansion of (1+x)^ n , n ∈ N, 0… JEE Main 2026 The coefficient of x^ 48 in (1+x)+2(1+x)^ 2 +3(1+x)^ 3 +…+100(1+x)^ 100 is equal to JEE Main 2026 Full Binomial Theorem list Binomial Theorem JEE Maths PYQs JEE Mathematics PYQs All JEE Main PYQs