sum and sigma notation: sum to n of 1/2(n)(n+1)

[Seimuna]

from the theorem,
sum of the first n positive integers = 1/2(n)(n+1)
sum of the square of the first n positive integers = 1/6(n)(n+1)(2n+1)

but wat if the question given not starting from 1 ??? can we stil apply the formula? if yes, how?

 

[Deleted member 4993]

Re: sum and sigma notation

from the theorem,
sum of the first n positive integers = 1/2(n)(n+1)
sum of the square of the first n positive integers = 1/6(n)(n+1)(2n+1)

but wat if the question given not starting from 1 ??? can we stil apply the formula? if yes, how?

If you were given a ruler where the first marking is 3" (instead of zero) - can you still use the ruler to measure length? If yes, how?

 

[Seimuna]

Re: sum and sigma notation

from the theorem,
sum of the first n positive integers = 1/2(n)(n+1)
sum of the square of the first n positive integers = 1/6(n)(n+1)(2n+1)

but wat if the question given not starting from 1 ??? can we stil apply the formula? if yes, how?

If you were given a ruler where the first marking is 3" (instead of zero) - can you still use the ruler to measure length? If yes, how?

minus it out? means if start from 3 to 10... then = (1/2)(10)(11) - 2 - 1 ? calculate it 1 by 1 ?

 

[Deleted member 4993]

Re: sum and sigma notation

\(\displaystyle \sum_{a_1}^{a_2}n^2 \, = \, \sum_{1}^{a_2}n^2 \, - \, \sum_{1}^{a_1-1}n^2 \,\)

 

[Seimuna]

Re: sum and sigma notation

\(\displaystyle \sum_{a_1}^{a_2}n^2 \, = \, \sum_{1}^{a_2}n^2 \, - \, \sum_{1}^{a_1-1}n^2 \,\)

thanks...