MHB -write expression in expanded form...find the sum

[karush]

nmh{2000} index{expanded form}
write each expression in expanded form and then find the sum
$
\begin{array}{l}
{{9}{\mathrm{)}}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\mathop{\sum}\limits_{{n}\mathrm{{=}}{3}}\limits^{5}{\mathrm{(}{n}^{2}}\mathrm{{-}}{2}^{n}{\mathrm{)}}}=(3^2-2^3)+(4^2-2^4)+(5^2-2^5)=-6
\\{{\mathrm{10}}{\mathrm{)}}\hspace{0.33em}\hspace{0.33em}\mathop{\sum}\limits_{{t}\mathrm{{=}}{1}}\limits^{5}{{t}{\mathrm{(}}{t}\mathrm{{-}}{1}{\mathrm{)}}}}
=1(1-1)+2(2-1)+3(3-1)+5(5-1)=40
\\{{\mathrm{11}}{\mathrm{)}}\hspace{0.33em}\hspace{0.33em}\mathop{\sum}\limits_{{i}\mathrm{{=}}{1}}\limits^{\mathrm{\infty}}{\mathrm{10}{\mathrm{\left({\frac{1}{2}}\right)}}^{i}}}
+10\left(\frac{1}{2}\right)^1
+10\left(\frac{1}{2}\right))^2
+10\left(\frac{1}{2}\right)^3
...=10\left(\frac{1}{2}\right)^i=10
\end{array}
$
hoped answers ok
no sure how to expand 11)

 

[Greg]

9. Correct.

10. You omitted the 4(4 - 1) term, otherwise correct.

11. It's an infinite geometric series that converges to a constant. Given that

$$\sum_{n=1}^{\infty}\left(\dfrac12\right)^{1/2}=\dfrac12\lim_{n\to\infty}\dfrac{1-r^n}{1-r}$$

with $r=\dfrac12$, can you compute the result?

Edit: Didn't see the '10' on the right. Ah well, no harm done. As for an expansion, what you have is sufficient I think, just omit the $10\left(\dfrac12\right)^i$ between the expansion and the '10'. ;)

 

[karush]

Okay gotit
Don't find these easy

Mahalo

BTW I tried math magic but wasn't worth it
However be nice to have the option
to render latex into an image if we want to use it someplace else

Cropping it out of post with gray background is painful