Sequences & Series
Telescoping sum of cubes
MJMT_Full_Test_11
Grade 12
Question:
$\dfrac{2^3-1^3}{1\times7}+\dfrac{4^3-3^3+2^3-1^3}{2\times11}+\dfrac{6^3-5^3+\cdots+1^3}{3\times15}+\cdots+\dfrac{30^3-29^3+\cdots+1^3}{15\times63}$ is equal to
Step-by-Step Solution
Key Concept: The $n$-th term numerator $=\sum_{k=1}^n[(2k)^3-(2k-1)^3]$. Show this simplifies to $n^2(4n+3)$. Denominator $=(2n-1)(4n+3)$. $T_n=n^2/(2n-1)$... simplify further to $T_n=n$.
Sum $=120$.
Correct Answer: 4