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

$$\sum _{n=2}^4(\frac{2n}{3}-1)=?$$

Answered by Frix last updated on 20/Apr/24

$$\mathrm{Seriously}?!$$$$...\mathrm{ok}:$$$$\underset{n=a}{\sum ^b}(c_{1}n+c_0)=\underset{n=a}{\sum ^b}\left(c_{1}n\right)+\underset{n=a}{\sum ^b}{c}_0=$$$$=\frac{c_1}{2}(a+b)(1+b-a)+c_0(1+b-a)=$$$$=\frac{((a+b)c_1+2c_0)(1+b-a)}{2}$$$$\mathrm{In}\mathrm{this}\mathrm{case}$$$$a=2\wedge b=4\wedge c_1=\frac{2}{3}\wedge c_0=-1$$$$\Rightarrow \mathrm{Answer}\mathrm{is}3$$$$$$$$...\mathrm{but}\mathrm{you}\mathrm{could}\mathrm{just}$$$$\frac{2\times 2}{3}-1+\frac{2\times 3}{3}-1+\frac{2\times 4}{3}-1=3$$

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

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

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

$$=\underset{n=2}{\sum ^4}\left(\frac{2n-3}{3}\right)=\frac{1}{3}\underset{n=2}{\sum ^4}(2n-3)$$$$=\frac{1}{3}\underset{n=1}{\sum ^3}((2n+1)-3)$$$$=\frac{1}{3}\underset{n=1}{\sum ^3}(2n-1)=\frac{1}{3}.\frac{3}{2}(1+5)$$$$=3$$$$$$

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