1 Find \( n \) if \( \sum_{a=3}^{5}(2 a+3 n)=51 \) 2 Prove that \( \sum_{n=1}^{p}(2 n-1) p=p^{3} \) 3 Find the value of \( p \) for which \( \sum_{x=1}^{p}(22-4 x)<-160 \) 4 Determine the largest value of \( n \) such that \[ \sum_{x=1}^{n}(3 x-2)<2000 \] 5 Calculate \( \sum_{r=5}^{12}(2 r-5) \sqrt{p} \) 6 What is the least value of \( n \) for which \( \sum_{p=1}^{n} 2 / 5(2)^{2 p-1}>11000 \)

To find \( n \) in the equation \( \sum_{a=3}^{5}(2 a+3 n)=51 \), we can start by evaluating the sum for \( a = 3, 4, 5 \). Plugging in these values gives us: - For \( a = 3 \): \( 2(3) + 3n = 6 + 3n \) - For \( a = 4 \): \( 2(4) + 3n = 8 + 3n \) - For \( a = 5 \): \( 2(5) + 3n = 10 + 3n \) Now, summing these results yields: \[ (6 + 3n) + (8 + 3n) + (10 + 3n) = 24 + 9n \] Setting this equal to 51 gives us: \[ 24 + 9n = 51 \Rightarrow 9n = 27 \Rightarrow n = 3 \] --- For the sum \( \sum_{n=1}^{p}(2 n-1) p=p^{3} \), we can first recognize that the left side simplifies to \( (1 + 3 + 5 + \ldots + (2p-1)) \times p \). The sum of the first \( p \) odd numbers is \( p^2 \). Therefore, the left side becomes \( p^2 \cdot p = p^3 \), which verifies the statement directly. --- To find \( p \) such that \( \sum_{x=1}^{p}(22-4 x)<-160 \), start by expressing the sum: \[ \sum_{x=1}^{p} (22 - 4x) = 22p - 4\sum_{x=1}^{p} x = 22p - 4 \frac{p(p+1)}{2} = 22p - 2p(p+1) \] Setting up the inequality: \[ 22p - 2p(p+1) < -160 \] This simplifies to: \[ 2p^2 - 2p + 160 > 0 \] Solving this quadratic equation will give the range for \( p \). --- The largest value of \( n \) such that \( \sum_{x=1}^{n}(3 x-2)<2000 \) can similarly be restructured. The sum calculates as: \[ \sum_{x=1}^{n}(3x - 2) = 3\sum_{x=1}^{n}x - 2n = 3 \frac{n(n+1)}{2} - 2n = \frac{3n(n+1)}{2} - 2n \] We need to solve \( \frac{3n(n+1)}{2} - 2n < 2000 \). --- To calculate \( \sum_{r=5}^{12}(2r-5) \sqrt{p} \), we simply evaluate: \[ \sum_{r=5}^{12} (2r-5) = (2(5) - 5) + (2(6) - 5) + ... + (2(12) - 5) \] Calculating each term yields an overall sum, which we multiply by \( \sqrt{p} \) for the final value. --- Lastly, for the condition \( \sum_{p=1}^{n} \frac{2}{5}(2)^{2p-1}>11000 \), notice that: \[ \sum_{p=1}^{n} \frac{2}{5} (2)^{2p-1} = \frac{2}{5} \left( 2^1 + 2^3 + 2^5 + ... + 2^{2n-1} \right) \] This forms a geometric series that can be summed up. Finding the smallest \( n \) by solving this inequality against 11000 will yield the desired result. Happy calculating! Each problem hides a little mathematical magic waiting to be unraveled. 🌟