need help on sequences topic

Find and simplify an expression for the sum of the natural numbers from (n + 1) to 2n inclusive.

i tried to solve it but couldnt i came to 0.5*(3n+1)(n-1) ...this is the answer according to my book (3/2)n^2 + n/2. could someone hint me towards getting this answer

Re: need help on sequences topic

Quote:

Originally Posted by

**abdulrehmanshah** Find and simplify an expression for the sum of the natural numbers from (n + 1) to 2n inclusive. according to my book (3/2)n^2 + n/2.

$\displaystyle \sum\limits_{k = n + 1}^{2n} k = \sum\limits_{k = 1}^{2n} k - \sum\limits_{k = 1}^n k $

Re: need help on sequences topic

edit: pipped at the post! (Giggle)

Re: need help on sequences topic

It is an AP with the first term = ( n+1), common difference 1 and last term = 2n

Also from n+1 to 2n we have n terms.

Using expression for sum of n terms of an AP we get S = n/2 [ 2a + ( n-1) d )

Thus we have (n+1) + ( n+2) + (n+3) + ..... + 2n = n/2 [ 2(n+1) + ( n-1) x 1 ] = n/2 [ 2n + 2 + n - 1 ] = n/2[ 3n + 1 ] = (3/2 ) n^2 + (n/2)

Re: need help on sequences topic

Quote:

Originally Posted by

**ibdutt** It is an AP with the first term = ( n+1), common difference 1 and last term = 2n

Also from n+1 to 2n we have n terms.

Using expression for sum of n terms of an AP we get S = n/2 [ 2a + ( n-1) d )

In order to do these problems you should know the so-called *Gauss Sum*:

$\displaystyle \sum\limits_{k = 1}^M k = \frac{{M\left( {M + 1} \right)}}{2}$.

So according to that we get $\displaystyle \frac{{2n(2n + 1)}}{2} - \frac{{n(n + 1)}}{2}$.