Let = ? r = 0 n ( 4 r 2 + 2 r + 1 ) C r n and = ( ? r = 0 n C r n r + 1 ) + 1 n + 1 . If 140

Question

Let $α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$ and $β = (\sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1}) + \frac{1}{n + 1} .$ If $140 < \frac{2 α}{β} < 281,$ then the value of $n$ is ________. [2024]

Smart Achievers · Accepted Answer

(5)

$α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$

$= 4 \sum_{r = 1}^{n} r^{2} \frac{n}{r} {}^{n - 1}{C_{r - 1}} + 2 \sum_{r = 1}^{n} r \frac{n}{r} {}^{n - 1}{C_{r - 1}} + \sum_{r = 0}^{n} {}^{n}{C_{r}}$

$= 4 n \sum_{r = 1}^{n} r \cdot {}^{n - 1}{C_{r - 1}} + 2 n \sum_{r = 1}^{n} {}^{n - 1}{C_{r - 1}} + \sum_{r = 0}^{n} {}^{n}{C_{r}}$

$= 4 n (n + 1) 2^{n - 2} + 2 n 2^{n - 1} + 2^{n} = 2^{n} [n (n + 1) + n + 1]$

$= 2^{n} [n^{2} + 2 n + 1] = 2^{n} {(n + 1)}^{2}$

Now, $β = \sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1} + \frac{1}{n + 1} = \sum_{r = 0}^{n} \frac{{}^{n + 1}{C_{r + 1}}}{n + 1} + \frac{1}{n + 1}$

$= \frac{1}{n + 1} \cdot 2^{n + 1} \Rightarrow \frac{2 α}{β} = 2 \cdot \frac{2^{n} {(n + 1)}^{2} \times (n + 1)}{2^{n + 1}} = {(n + 1)}^{3}$

Given that, $140 < \frac{2 α}{β} < 281 \Rightarrow 140 < {(n + 1)}^{3} < 281$

Now, for $n = 4; {(n + 1)}^{3} = 5^{3} = 125$

for $n = 5; {(n + 1)}^{3} = 6^{3} = 216$

for $n = 6; {(n + 1)}^{3} = 7^{3} = 343$

Hence, the value of $n = 5$

Solution

Q.
Let $α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$ and $β = (\sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1}) + \frac{1}{n + 1} .$ If $140 < \frac{2 α}{β} < 281,$ then the value of $n$ is ________. [2024]

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Solution

Q. Let α=∑r=0n(4r2+2r+1)Crn and β=(∑r=0nCrnr+1)+1n+1. If 140<2αβ<281, then the value of n is ________. [2024]

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Q.
Let $α = \sum_{r = 0}^{n} (4 r^{2} + 2 r + 1) {}^{n}{C_{r}}$ and $β = (\sum_{r = 0}^{n} \frac{{}^{n}{C_{r}}}{r + 1}) + \frac{1}{n + 1} .$ If $140 < \frac{2 α}{β} < 281,$ then the value of $n$ is ________. [2024]