Question and Answers Forum

All Questions      Topic List

Algebra Questions

Previous in All Question      Next in All Question      

Previous in Algebra      Next in Algebra      

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

Σ_(k=6) ^(n+6) (k−4)=((an^2 +bn+c)/2) a+b+c=?

$$\underset{{k}=\mathrm{6}} {\overset{{n}+\mathrm{6}} {\sum}}\left({k}−\mathrm{4}\right)=\frac{{an}^{\mathrm{2}} +{bn}+{c}}{\mathrm{2}} \\ $$$${a}+{b}+{c}=? \\ $$

Commented by bobhans last updated on 30/Jan/22

Σ_(k=6) ^(n+6) (k−4)=Σ_(k=1) ^(n+1) (k+1)=Σ_(k=1) ^(n+1) k+Σ_(k=1) ^(n+1) 1 = (((n+1)(n+2))/2)+(n+1) =((n^2 +3n+2+2n+2)/2) = ((n^2 +5n+4)/2) ⇒a+b+c = 1+5+4=10

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

Answered by mahdipoor last updated on 26/Jan/22

=(6+7+...+(n+6))−(4+...+4)= ((((n+6)(n+7))/2)−((6×5)/2))−4(n+1)= ((n^2 +5n−38)/2)≡((an^2 +bn+c)/2) ⇒a+b+c=1+5−38=−32

$$=\left(\mathrm{6}+\mathrm{7}+...+\left({n}+\mathrm{6}\right)\right)−\left(\mathrm{4}+...+\mathrm{4}\right)= \\ $$$$\left(\frac{\left({n}+\mathrm{6}\right)\left({n}+\mathrm{7}\right)}{\mathrm{2}}−\frac{\mathrm{6}×\mathrm{5}}{\mathrm{2}}\right)−\mathrm{4}\left({n}+\mathrm{1}\right)= \\ $$$$\frac{{n}^{\mathrm{2}} +\mathrm{5}{n}−\mathrm{38}}{\mathrm{2}}\equiv\frac{{an}^{\mathrm{2}} +{bn}+{c}}{\mathrm{2}} \\ $$$$\Rightarrow{a}+{b}+{c}=\mathrm{1}+\mathrm{5}−\mathrm{38}=−\mathrm{32} \\ $$

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

2+3+4+...+(n+2) =(((n+1)(n+4))/2) =((n^2 +5n+4)/2) a+b+c=1+5+4=10

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com