yummydum
Using sigma notation, describe the total length of 7 pictures hung side-by-side. Then find the total length the pictures occupy along the wall if the first picture has a length of 10 inches and the length of each successive picture is 4 inches longer than the previous one.
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yummydum
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@hartnn
hartnn
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let l1 to l7 be length of pics.
from n=1 to 7, l_n and sum notation :
\(\sum \limits_{n=1}^7l_n\)
is your 1st part.
ParthKohli
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Hint: for the first part, denote the length of the seven pictures by a sequence \(a_1, a_2 \cdots a_7\)
hartnn
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l1 =10 and common difference d =4
yummydum
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rightI got that but for the equation..?
hartnn
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you need to find n'th tern l_n
with l_1 =10 , d = 4
\(l_n = l_1+(n-1)d\)
yummydum
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10, 14, 18,..\[a_n=(10)-(n-1)(4)\]\[a_n=10-4n+4\]\[a_n=14-4n\]\[\sum_{n=1}^{7}~14n-4\]yes?
hartnn
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middle sign is +
ln = 6+4n
yummydum
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oh ...i knew that..
hartnn
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now can you solve the summation ?
hartnn
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\(\sum 6+4\sum n\)
ParthKohli
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The worst pain, after the famines and diseases, is typing on an iPad. :-(
hartnn
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know whats \(\sum n=..?\)
yummydum
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wow it wrote that weird..
\[\large\sum_{n=1}^{7} ~6+4n~~ \implies~~154\]
ParthKohli
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I don't know, just pointing out that a common mistake students sometimes make is that they do \(\sum 6 = 6\) when it's not.
hartnn
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i also get 154
ParthKohli
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You didn't make that mistake though, nice :-)
yummydum
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err..what?
i did \([6+(4\times1)]+[6+(4\times2)]+[6+(4\times3)]+\)\([6+(4\times4)]+[6+(4\times5)]+[6+(4\times6)]+[6+(4\times7)]\)
ParthKohli
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You are right.
yummydum
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10+14+18+22+26+30+34=154.....yup :)
ParthKohli
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I never said you were wrong lol, just pointing out a common mistake.
ParthKohli
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Good stuff :-)
yummydum
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lol, thanks guys!
yummydum
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whats 8(n+3)??
yummydum
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10+14+18+22+26+30+34=154 ....is that not right?
yummydum
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o.O
yummydum
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lol no probs