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c1 question

Evaluate

30 r=10∑ (7 + 2r)

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

∑1030(7+2r)=∑130(7+2r)−∑19(7+2r)\sum_{10}^{30} (7+2r) =\sum_{1}^{30} (7+2r) -\sum_{1}^{9} (7+2r)
Reply 2
Avatar for Mina_
Mina_
OP
8
Original post by NotNotBatman
split it so it's like this;

∑1030(7+2r)=∑130(7+2r)−∑19(7+2r)\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
since when is this c1??, what board are you doing ?
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


Sn=n2(a+l)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 r=10 to r=30 r=30 then that's

Unparseable latex formula:

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 r=10 to r=30 r = 30 is the same as summing from r=1 r=1 to r=30r=30 subtract the sum form r=1 r =1 to r=9r=9

because
Unparseable latex formula:

\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 ∑19=u1+u2+u3+u4+u5+u6+u7+u8+u9\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
Unparseable latex formula:

u_1_0

to
Unparseable latex formula:

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).
Reply 5
Avatar for Mina_
Mina_
OP
8
Original post by NotNotBatman
Sn=n2(a+l)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 r=10 to r=30 r=30 then that's

Unparseable latex formula:

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 r=10 to r=30 r = 30 is the same as summing from r=1 r=1 to r=30r=30 subtract the sum form r=1 r =1 to r=9r=9

because
Unparseable latex formula:

\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 ∑19=u1+u2+u3+u4+u5+u6+u7+u8+u9\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
Unparseable latex formula:

u_1_0

to
Unparseable latex formula:

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!
Reply 6
Avatar for Mina_
Mina_
OP
8
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
Reply 7
Avatar for JN17
JN17
10
Just sub in 10 and 30, your 'n' value is 30-10+1= 21, so 21/2(a+L)
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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