Mina_
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#1
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#1
Evaluate

30 r=10∑ (7 + 2r)

30 is above the sigma sign and r=10 is below
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NotNotBatman
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#2
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#2
split it so it's like this;

\sum_{10}^{30} (7+2r) =\sum_{1}^{30} (7+2r) -\sum_{1}^{9} (7+2r)
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Mina_
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#3
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#3
(Original post by NotNotBatman)
split it so it's like this;

\sum_{10}^{30} (7+2r) =\sum_{1}^{30} (7+2r) -\sum_{1}^{9} (7+2r)
can you explain it further and what you'd do?
the mark scheme uses n/2 ( a + L) i believe which isnt what we've learnt
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Science_help
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#4
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#4
since when is this c1??, what board are you doing ?
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NotNotBatman
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#5
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#5
(Original post by Mina_)
can you explain it further and what you'd do?
the mark scheme uses n/2 ( a + L) i believe which isnt what we've learnt
S_n =\frac{n}{2} (a+l) where a is the first term and l is the last term, it's a standard formula which you should learn.

So if you have to sum from  r=10 to  r=30 then that's

 U_1_0 + U_11 + U_1_2 + ... + U_3_0,

but this formula uses the first term, so you have to realise that the sum from  r=10 to  r = 30 is the same as summing from  r=1 to r=30 subtract the sum form  r =1 to r=9

because  \sum_{1}^{30}= u_1 + u_2 + u_3 +u_4 + u_5 + u_6 + u_7 + u_8 + u_9 + u_1_0+ u_1_1+...+u_3_0

and \sum_{1}^{9} = u_1 + u_2 + u_3 +u_4 + u_5 + u_6 + u_7 + u_8 + u_9

and notice when you subtract them you're left with the summation form  u_1_0 to  u_3_0
which is what you're asked for, so apply the summation formula remembering a is the first term (the number at the bottom) and l is the last term (the number at the top).
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Mina_
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#6
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#6
(Original post by NotNotBatman)
S_n =\frac{n}{2} (a+l) where a is the first term and l is the last term, it's a standard formula which you should learn.

So if you have to sum from  r=10 to  r=30 then that's

 U_1_0 + U_11 + U_1_2 + ... + U_3_0,

but this formula uses the first term, so you have to realise that the sum from  r=10 to  r = 30 is the same as summing from  r=1 to r=30 subtract the sum form  r =1 to r=9

because  \sum_{1}^{30}= u_1 + u_2 + u_3 +u_4 + u_5 + u_6 + u_7 + u_8 + u_9 + u_1_0+ u_1_1+...+u_3_0

and \sum_{1}^{9} = u_1 + u_2 + u_3 +u_4 + u_5 + u_6 + u_7 + u_8 + u_9

and notice when you subtract them you're left with the summation form  u_1_0 to  u_3_0
which is what you're asked for, so apply the summation formula remembering a is the first term (the number at the bottom) and l is the last term (the number at the top).
Thank you!
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Mina_
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#7
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#7
(Original post by Science_help)
since when is this c1??, what board are you doing ?
Edexcel, haha this is from the solomon papers which is typically harder
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JN17
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#8
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Just sub in 10 and 30, your 'n' value is 30-10+1= 21, so 21/2(a+L)
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Science_help
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#9
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#9
(Original post by Mina_)
Edexcel, haha this is from the solomon papers which is typically harder
ahh adexcel, makes sense, in aqa we have this topic in c2, similar to how circle geometry is on c2 for edexcel, whereas its c1 for aqa?
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