Question 696783
{{{sum(n^2, n=0, k)= k(k+1)(2k+1)/6}}}
 
For k=1:
{{{sum(n^2, n=0, k)=sum(n^2, n=0, 1)= 0^2+1^2=0+1=1}}}
{{{k(k+1)(2k+1)/6=1*(1+1)*(2*1+1)/6=1*2*3/6=6/6=1}}}
So the formula is valid for k=1.
 
For k=2:
{{{sum(n^2, n=0, k)= sum(n^2, n=0, 2)= 0^2+1^2+2^2=0+1+4=5}}}
{{{k(k+1)(2k+1)/6=2*(2+1)*(2*2+1)/6=2*3*5/6=30/6=5}}}
So the formula is valid for k=2.
 
For k=3:
{{{sum(n^2, n=0, k)= sum(n^2, n=0, 3)= 0^2+1^2+2^2+3^2=0+1+4+9=14}}}
{{{k(k+1)(2k+1)/6=3*(3+1)*(2*3+1)/6=3*4*7/6=84/6=14}}}
So the formula is valid for k=3.
 
BIG SUM COMPUTATION (step by step):
{{{sum((2m^2+3m-4), m=0, 5)=sum(2m^2, m=0, 5)+sum(3m, m=0, 5)+sum(-4, m=0, 5)}}}
(Changing the order and grouping in the sum, just like you do when you add polynomials and "collect like terms").
{{{sum((2m^2+3m-4), m=0, 5)=sum(2m^2, m=0, 5)+sum(3m, m=0, 5)+sum(-4, m=0, 5)=2*sum(m^2, m=0, 5)+3*sum(m, m=0, 5)-4*sum(1, m=0, 5)}}}
(taking out common factors as the next step)
{{{sum((2m^2+3m-4), m=0, 5)=sum(2m^2, m=0, 5)+sum(3m, m=0, 5)+sum(-4, m=0, 5)=2*sum(m^2, m=0, 5)+3*sum(m, m=0, 5)-4*sum(1, m=0, 5)=5(5+1)(2*5+1)/6+3*15-4*6}}}
(applying the formula given and easy to calculate values for the second and third sums).
Now we finish the indicated calculations:
{{{sum((2m^2+3m-4), m=0, 5)=sum(2m^2, m=0, 5)+sum(3m, m=0, 5)+sum(-4, m=0, 5)=2*sum(m^2, m=0, 5)+3*sum(m, m=0, 5)-4*sum(1, m=0, 5)=2*(5(5+1)(2*5+1)/6)+3*15-4*6=2*55+45-24=110+45-24=highlight(131)}}}
 
KNOWN SUMS:
You know that {{{sum(1, m=0, 5)=1+1+1+1+1+1=6}}} because it is a sum of six terms (m=0 to m=5) and all the terms are 1.
The calculation of {{{sum(m, m=0, 5)=0+1+2+3+4+5}}} is also easy.
You know that is an arithmetic series.
You can calculate it by adding the terms I already listed above.
{{{sum(m, m=0, 5)=0+1+2+3+4+5=15}}}
Otherwise you can apply the formula for sum of an arithmetic sequence
{{{sum(a[m], m=0, k)=(a[0]+a[k])*(k+1)/2}}} and get {{{sum(a[m], m=0, 5)=(0+5)*(5+1)/2=5*6/2=15}}}