n-2-4-2n-3-1-

Question Number 206622 by necx122 last updated on 20/Apr/24

$$\sum_{{n}=\mathrm{2}} ^{\mathrm{4}} \left(\frac{\mathrm{2}{n}}{\mathrm{3}}\:−\mathrm{1}\right)\:=\:? \\ $$

Answered by Frix last updated on 20/Apr/24

Seriously?! ...ok: Σ_(n=a) ^b (c_1 n+c_0 )=Σ_(n=a) ^b (c_1 n)+Σ_(n=a) ^b c_0 = =(c_1 /2)(a+b)(1+b−a)+c_0 (1+b−a)= =((((a+b)c_1 +2c_0 )(1+b−a))/2) In this case a=2∧b=4∧c_1 =(2/3)∧c_0 =−1 ⇒ Answer is 3 ...but you could just ((2×2)/3)−1+((2×3)/3)−1+((2×4)/3)−1=3

$$\mathrm{Seriously}?! \\ $$$$…\mathrm{ok}: \\ $$$$\underset{{n}={a}} {\overset{{b}} {\sum}}\left({c}_{\mathrm{1}} {n}+{c}_{\mathrm{0}} \right)=\underset{{n}={a}} {\overset{{b}} {\sum}}\left({c}_{\mathrm{1}} {n}\right)+\underset{{n}={a}} {\overset{{b}} {\sum}}{c}_{\mathrm{0}} = \\ $$$$=\frac{{c}_{\mathrm{1}} }{\mathrm{2}}\left({a}+{b}\right)\left(\mathrm{1}+{b}−{a}\right)+{c}_{\mathrm{0}} \left(\mathrm{1}+{b}−{a}\right)= \\ $$$$=\frac{\left(\left({a}+{b}\right){c}_{\mathrm{1}} +\mathrm{2}{c}_{\mathrm{0}} \right)\left(\mathrm{1}+{b}−{a}\right)}{\mathrm{2}} \\ $$$$\mathrm{In}\:\mathrm{this}\:\mathrm{case} \\ $$$${a}=\mathrm{2}\wedge{b}=\mathrm{4}\wedge{c}_{\mathrm{1}} =\frac{\mathrm{2}}{\mathrm{3}}\wedge{c}_{\mathrm{0}} =−\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{Answer}\:\mathrm{is}\:\mathrm{3} \\ $$$$ \\ $$$$…\mathrm{but}\:\mathrm{you}\:\mathrm{could}\:\mathrm{just} \\ $$$$\frac{\mathrm{2}×\mathrm{2}}{\mathrm{3}}−\mathrm{1}+\frac{\mathrm{2}×\mathrm{3}}{\mathrm{3}}−\mathrm{1}+\frac{\mathrm{2}×\mathrm{4}}{\mathrm{3}}−\mathrm{1}=\mathrm{3} \\ $$

Commented by necx122 last updated on 20/Apr/24

Thank yiubso much much sir Frix. Posted the question because I was shocked about how I solved the question, got 3 but couldn't see it as an option. I suspect that they made a typo error in the question I used. The approach is same with what you did. Thank you so much sir.

Commented by Frix last updated on 20/Apr/24

You′re welcome. Sorry but sometimes I′m shocked when people who already asked for integrals all of a sudden ask for things like 1+1=?

$$\mathrm{You}'\mathrm{re}\:\mathrm{welcome}. \\ $$$$\mathrm{Sorry}\:\mathrm{but}\:\mathrm{sometimes}\:\mathrm{I}'\mathrm{m}\:\mathrm{shocked}\:\mathrm{when} \\ $$$$\mathrm{people}\:\mathrm{who}\:\mathrm{already}\:\mathrm{asked}\:\mathrm{for}\:\mathrm{integrals}\:\mathrm{all} \\ $$$$\mathrm{of}\:\mathrm{a}\:\mathrm{sudden}\:\mathrm{ask}\:\mathrm{for}\:\mathrm{things}\:\mathrm{like}\:\mathrm{1}+\mathrm{1}=? \\ $$

Commented by necx122 last updated on 20/Apr/24

Trust me, I can relate: it's really funny. At times, I just need to be sure that I've not started solving wrongly the very easy math problems while trying do the tedious stuff.

Answered by cortano21 last updated on 21/Apr/24

= Σ_(n= 2) ^4 (((2n−3)/3)) = (1/3)Σ_(n= 2) ^4 (2n−3) = (1/3) Σ_(n=1) ^3 ((2n+1)−3) = (1/3) Σ_(n=1) ^3 (2n−1)=(1/3).(3/2)(1+5) = 3

$$\:\:=\:\underset{{n}=\:\mathrm{2}} {\overset{\mathrm{4}} {\sum}}\left(\frac{\mathrm{2}{n}−\mathrm{3}}{\mathrm{3}}\right)\:=\:\frac{\mathrm{1}}{\mathrm{3}}\underset{{n}=\:\mathrm{2}} {\overset{\mathrm{4}} {\sum}}\left(\mathrm{2}{n}−\mathrm{3}\right) \\ $$$$\:\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\left(\left(\mathrm{2}{n}+\mathrm{1}\right)−\mathrm{3}\right) \\ $$$$\:=\:\frac{\mathrm{1}}{\mathrm{3}}\:\underset{{n}=\mathrm{1}} {\overset{\mathrm{3}} {\sum}}\left(\mathrm{2}{n}−\mathrm{1}\right)=\frac{\mathrm{1}}{\mathrm{3}}.\frac{\mathrm{3}}{\mathrm{2}}\left(\mathrm{1}+\mathrm{5}\right) \\ $$$$\:\:=\:\mathrm{3} \\ $$$$ \\ $$

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