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

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

Commented by bobhans last updated on 30/Jan/22

$$\underset{\mathrm{k}=6}{\sum ^{\mathrm{n}+6}}(\mathrm{k}-4)=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}+1}}(\mathrm{k}+1)=\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}+1}}\mathrm{k}+\underset{\mathrm{k}=1}{\sum ^{\mathrm{n}+1}}1$$$$=\frac{(\mathrm{n}+1)(\mathrm{n}+2)}{2}+(\mathrm{n}+1)$$$$=\frac{{\mathrm{n}}^2+3\mathrm{n}+2+2\mathrm{n}+2}{2}=\frac{{\mathrm{n}}^2+5\mathrm{n}+4}{2}$$$$\Rightarrow \mathrm{a}+\mathrm{b}+\mathrm{c}=1+5+4=10$$

Answered by mahdipoor last updated on 26/Jan/22

$$=(6+7+...+(n+6))-(4+...+4)=$$$$(\frac{(n+6)(n+7)}{2}-\frac{6\times 5}{2})-4(n+1)=$$$$\frac{n^2+5n-38}{2}\equiv \frac{{an}^2+bn+c}{2}$$$$\Rightarrow a+b+c=1+5-38=-32$$

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

$$2+3+4+...+(n+2)$$$$=\frac{(n+1)(n+4)}{2}$$$$=\frac{n^2+5n+4}{2}$$$$a+b+c=1+5+4=10$$

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