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sarah12345
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#1
find an expression for the sum of natural numbers from (n+1) to 2n inclusive

heres what i did

s=n/2 (2n+2+2n-1) = n/2(4n+1)

can you see where i went wrong?

thanks
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Gaz031
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Consider the sum to 2n - the sum to n.
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lgs98jonee
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(Original post by Gaz031)
Consider the sum to 2n - the sum to n.
yeh

numbers up to 2n are

1,2,3,4...n,n+1,n+2....2n

so you want all the ones after n, so the way to do that is take the sum from 1 to 2n and subtract from it the ones from 1 to n like this

sum=2n(2n+1)/2-n(n+1)/2
=n/2*(4n+2-n-1)
=n(3n+1)/2
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john !!
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(Original post by sarah12345)
find an expression for the sum of natural numbers from (n+1) to 2n inclusive

heres what i did

s=n/2 (2n+2+2n-1) = n/2(4n+1)

can you see where i went wrong?

thanks
you have

Code:
Sum =      (n+1)  + (n+2)  + (n+3)   + ... + (2n-3) + (2n-2) + (2n-1) + 2n
Sum = 2n + (2n-1) + (2n-2) + (2n-3) + ...  + (n+3)  + (n+2)  + (n+1)
Each sum has n terms. (and excluding the 2n at the end/start, n-1 terms)

Adding these,

2*Sum = 2n + 3n + 3n + 3n + ... + 3n + 3n + 3n + 2n
2*Sum = 4n + (n-1)(3n)
2*Sum = n + 3n^2
Sum = n(3n + 1)/2
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Newton
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A. P.=(n+1), (n+2), (n+3), (n+4),..., (2n-2), (2n-1), 2n

Sn=(n+1)+(n+2)+(n+3)+(n+4)+...+( 2n-2)+(2n-1)+2n

Reverse:

Sn=2n+(2n-1)+(2n-2)+...+(n+4)+(n+3)+(n+2)+(n+1)

Adding respective terms:

2Sn=(3n+1)+(3n+1)+...+(3n+1)

Therefore:

Sn=(n/2)(3n+1)

Newton.
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Christophicus
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Or a much simpler way.
Sn = 0.5n(a + L)
Sn = 0.5n(n + 1 + 2n)
Sn = 0.5n(3n + 1)
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VCVT17
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#7
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I think you must find the sum of nature numbers from 1 to 2n first and then subtraction to the sum of nature number from 1 to n.
S(2n) - S(n) = n*(3n+1)
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