Question

find the sum :
Σ(r=1 to n-1) {[r+1]^3-1}

Answer 1

*[(r+1)^3 - 1 ]

Answer 2

1+1^3-1

Answer 3

2^3-1

Answer 4

3-1 = 2

Answer 5

2^2 = ?

Answer 6

2^2 = 2 times 2 = ?

Answer 7

huh

Answer 8

answer is in form of n

Answer 9

Hint:
we have:
\[{\left( {r + 1} \right)^3} - 1 = {r^3} + 3{r^2} + 3r\]
so we can split your sum as follows:
\[\sum\limits_1^{n - 1} {\left[ {{{\left( {r + 1} \right)}^3} - 1} \right]} = \sum\limits_1^{n - 1} {{r^3}} + 3\sum\limits_1^{n - 1} {{r^2}} + 3\sum\limits_1^{n - 1} r \]

Answer 10

this one one will time taking
there is one way to simplify that into r^3 -1 only