Example ① Express Σ_{k=1}^{n} k(k+1) in terms of n.
Solution:Σ(k²+k) = n(n+1)(2n+1)/6 + n(n+1)/2
= n(n+1){(2n+1)+3}/6 = n(n+1)(2n+4)/6 = n(n+1)(n+2)/3
Solution:Σ(k²+k) = n(n+1)(2n+1)/6 + n(n+1)/2
= n(n+1){(2n+1)+3}/6 = n(n+1)(2n+4)/6 = n(n+1)(n+2)/3
Key Point
- Factoring out the common factor n(n+1) makes the expression simpler
Practice Problems
- Express Σ_{k=1}^{n} (2k−1) in terms of n.
Solutions and Explanations
- Solution:n²
2·n(n+1)/2 − n = n². The sum of the odd numbers.