Thread: Problem With A Sequence Proof

The theorem for finding the sum of n terms of an Arithmetic Sequence is:

Sn = n/2[ 2a + (n - 1)d ] = n/2(a + an)


However, I am having a problem with the proof of this which is given in the book. I will show the proof given and stop at the point where I'm having a problem understanding.

Sn = a1 + a2 + a3 + ... + an ;...........................sum of first n terms

= a + (a + d) + (a + 2d) + ... + [a + (n - 1)d] ;....uses an = a + (n - 1)d

= (a + a + ... + a) + [d + 2d + ... + (n - 1)d] ;.....rearranging terms

= na + d[1 + 2 + ... + (n - 1)] ;.........................there are n a terms and factors out d

= na + d[ (n - 1)n / 2 ] ;..................................inside the brackets using property 1 + 2 + 3 + ... + n = n(n+1)/2


It is at this point in the proof that I have a problem, since n(n-1)/2 is not the same as n(n+1)/2.

Originally Posted by spiritualfields

...
= na + d[1 + 2 + ... + (n - 1)] ;.........................there are n a terms and factors out d

= na + d[ (n - 1)n / 2 ] ;..................................inside the brackets using property 1 + 2 + 3 + ... + n = n(n+1)/2

It is at this point in the proof that I have a problem, since n(n-1)/2 is not the same as n(n+1)/2.

Hello,

you are right the sum of the first n summands is:

s_n = n/2*(n-1)

You don't have n summands but (n-1) summands. Use the formula to calculate the sum of the first (n-1) summands:

s_(n-1) = (n-1)/2*((n-1) + 1) = (n-1)/2 * n = n(n-1)/2

EB

Earboth

s_(n-1) = (n-1)/2*((n-1) + 1) = (n-1)/2 * n = n(n-1)/2

Yes, now I can see how substituting (n - 1) for n causes the sign change. Thanks for the help.