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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

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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#3

(Original post by

Consider the sum to 2n - the sum to n.

**Gaz031**)Consider the sum to 2n - the sum to n.

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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#4

(Original post by

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

**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

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)

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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#5

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.

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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#7

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)

S(2n) - S(n) = n*(3n+1)

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