k-6-n-6-k-4-an-2-bn-c-2-a-b-c-

Question Number 165157 by mathlove last updated on 26/Jan/22

\underset{k = 6}{\sum^{n + 6}} (k - 4) = \frac{{a n}^{2} + b n + c}{2}

a + b + c = ?

Commented by bobhans last updated on 30/Jan/22

\underset{k = 6}{\sum^{n + 6}} (k - 4) = \underset{k = 1}{\sum^{n + 1}} (k + 1) = \underset{k = 1}{\sum^{n + 1}} k + \underset{k = 1}{\sum^{n + 1}} 1

= \frac{(n + 1) (n + 2)}{2} + (n + 1)

= \frac{n^{2} + 3 n + 2 + 2 n + 2}{2} = \frac{n^{2} + 5 n + 4}{2}

\Rightarrow a + b + c = 1 + 5 + 4 = 10

Answered by mahdipoor last updated on 26/Jan/22

= (6 + 7 + \dots + (n + 6)) - (4 + \dots + 4) =

(\frac{(n + 6) (n + 7)}{2} - \frac{6 \times 5}{2}) - 4 (n + 1) =

\frac{n^{2} + 5 n - 38}{2} \equiv \frac{{a n}^{2} + b n + c}{2}

\Rightarrow a + b + c = 1 + 5 - 38 = - 32

Answered by mr W last updated on 30/Jan/22

2 + 3 + 4 + \dots + (n + 2)

= \frac{(n + 1) (n + 4)}{2}

= \frac{n^{2} + 5 n + 4}{2}

a + b + c = 1 + 5 + 4 = 10

Facebook Tweet Pin