Question

$\sum _{k=1}^5\frac{1^3+2^3+....+k^3}{1+3+5+.....+\left(2k-1\right)}=$

• A. 22.5
• B. 24.5
• C. 28.5
• D. 32.5

Answer 1

Answer: (A)

$\sum _{k=1}^5\frac{(k+1{)}^2}{4}=\frac{4}{4}+\frac{9}{4}+\frac{16}{4}+\frac{25}{4}+\frac{36}{4}=\frac{90}{4}=22.5$