how do i make the sum of [3+4+5+...+(n+1)] into 1/2[ n-1)(n+4)]??
isn't it only 1/2(n(n+4))? Sn=n/2(a+l)?
Thx.
Originally posted by Bigcable22:how do i make the sum of [3+4+5+...+(n+1)] into 1/2[ n-1)(n+4)]??
isn't it only 1/2(n(n+4))? Sn=n/2(a+l)?
Thx.
Hi Bigcable22,
Adding the first number to the last number in the series, you will have ( n + 4 ).
( n - 1 ) is the amount of numbers in the series.
I believe the formula will be something similar to [ First number + Last Number ] X (Total amount of numbers)/2
Thus, the sum of the series will be ( n + 4 ) X ( n - 1 )/2 = 1/2( n - 1 )( n + 4 ).
Cheers.