Question 5 Prove that for any positive integer nn, ∑k=1n(3k+2)=3n(n+1)2+2n.\sum_{k=1}^{n} \bigl(3k+2\bigr) = \frac{3n(n+1)}{2}+2n. Show solutionHide solution+Question 5 - Solution We prove the identity using properties of summations. Start with the given sum: ∑k=1n(3k+2).\sum_{k=1}^{n} (3k+2). Split the sum using linearity: ∑k=1n(3k+2)=∑k=1n3k+∑k=1n2.\sum_{k=1}^{n} (3k+2) = \sum_{k=1}^{n} 3k + \sum_{k=1}^{n} 2. Factor out constants: ∑k=1n3k=3∑k=1nk,∑k=1n2=2∑k=1n1.\sum_{k=1}^{n} 3k = 3\sum_{k=1}^{n} k, \qquad \sum_{k=1}^{n} 2 = 2\sum_{k=1}^{n} 1. Use known summation formulas: ∑k=1nk=n(n+1)2,∑k=1n1=n.\sum_{k=1}^{n} k=\frac{n(n+1)}{2}, \qquad \sum_{k=1}^{n} 1=n. Substitute these results: ∑k=1n(3k+2)=3⋅n(n+1)2+2n.\sum_{k=1}^{n} (3k+2) = 3\cdot\frac{n(n+1)}{2} + 2n. Thus, ∑k=1n(3k+2)=3n(n+1)2+2n.\sum_{k=1}^{n} (3k+2) = \frac{3n(n+1)}{2}+2n. ∑k=1n(3k+2)=3n(n+1)2+2n\boxed{\sum_{k=1}^{n} (3k+2)=\frac{3n(n+1)}{2}+2n}
