Question

Find $\displaystyle\sum_{r=1}^{n} (6r^{2}+4r-3)$

Answer 1

Answer:

n²(2n + 5)

Step-by-step explanation:

∑ (6r² + 4r -3)

= ∑ 6r² + ∑4r  - ∑3

= 6 n(n + 1)(2n + 1)/6  + 4n(n+1)/2 - 3n

= n(n + 1)(2n + 1) + 2n(n + 1) - 3n

= n(n + 1) (2n + 1  + 2) - 3n

= n(n + 1) (2n + 3) - 3n

= n (n + 1) (2n + 3) - 3)

= n ( 2n² + 5n + 3 - 3)

= n(2n² + 5n)

= n²(2n + 5)

∑ (6r² + 4r -3) = n²(2n + 5)

Answer 2

Answer:

n²(2n + 5)

Step-by-step explanation:

In this question,

We need to find the $\sum(6r^{2}\ +\ 4r\ -\ 3)$

∑ (6r² + 4r - 3)

= ∑ 6r² + ∑4r  - ∑3

= $\frac{6n(n\ +\ 1)(2n\ +\ 1)}{6}\ +\ \frac{4n(n+1)}{2}\ -\ 3n$

= n(n + 1)(2n + 1) + 2n(n + 1) - 3n

= n(n + 1) (2n + 1  + 2) - 3n

= n(n + 1) (2n + 3) - 3n

= n (n + 1) (2n + 3) - 3)

= n ( 2n² + 5n + 3 - 3)

= n(2n² + 5n)

= n²(2n + 5)

∑ (6r² + 4r -3) = n²(2n + 5)