We conjecture that there is a formula of the form $\sum_{j=1}^{n}{j^2} = an^3 + bn^2 + cn + d$ for all integers n ≥ 1 (3)
(a) Assuming that such a formula is true, find the value of a, b, c, d. (Hint: Substituting the first few values of n yields a system of linear equations in a, b, c, d, which has a unique solution).
So for this I have that
1 = a + b + c + d
4 = 8a + 4b + 2c + d
9 = 27a + 9b + 3c + d
16 = 64a + 16b + 4c + d
Which after I reduced it in a matrix yields a=0, b=1 c=0 d=0; Have I done this part correctly?
(b) With the constants a, b, c, d you found in part (a), use mathematical induction to prove (3).
So now I have ∑j=1 to n such that j^2 = n^2
Base case:
for n = 1 LHS = 1 = RHS = 1
Assume true for n=k ∑j=1 to k such that j^2 = k^2
Prove true for n= k+1 ∑j=1 to k+1 such that j^2 = ∑j=1 to k such that j^2 + (k+1)^2
= k^2 + (k+1)^2
I have no idea where to go from here though.
"(Hint: Substituting the first few values of n yields a system of linear equations in a, b, c, d, which has a unique solution)."
So
$\sum_{j=0}^0 j^2 = 0 = a*0^3 + b*0^2 + c*0 + d = d$
$\sum_{j=0}^1 j^2 = 1 = a*1^3 + b*1^2 + c*1 + d = a + b + c +d$
$\sum_{j=0}^2 j^2 = 1 + 4 = 5 = a*2^3 + b*2^2 + c*2 + d = 8a + 4 b + 2c + d$
$\sum_{j=0}^3 j^2 = 1 + 4 + 9= 14 = 27a + 9b + 3c + d$
So $d = 0$. $a = 1/3, b = 1/2, c = 1/6$.
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Prove by induction:
In solving the linear equations we've shown it is true for $n = 0,1,2,3$
Assume it is true for $n = k$. Then:
$\sum_{j=1}^{k+1} j^2 = \sum_{j=1}^k j^2 + (k + 1)^2=$
$k^3*1/3 + k^2*1/2 + k*1/6 + (k^2 + 2k + 1)=$
$k^3* 1/3 + k^2(3/2) + k(7/6) + 1$.
Now
$(k+1)^3*1/3 + (k+1)^2*1/2 + (k+1)*1/6 = $
$(k^3 + 3k^2 + 3k+ 1)*1/3 + (k^2 + 2k + 1)*1/2 + k/6 + 1/6 = $
$k^3*1/3 + k^2 + k + 1/3 + k^2/2 + k + 1/2 + k/6 + 1/6 = $
$k^3 *1/3 + k^2(3/2) + k(7/6) + 1/3 + 1/2 + 1/6 = $
$k^3 *1/3 + k^2(3/2) + k(7/6) +1 = $
So $\sum_{j=1}^{k+1} j^2 =(k+1)^3*1/3 + (k+1)^2*1/2 + (k+1)*1/6 $
So by induction, we have shown $\sum_{j=0}^n j^2 = 1/3*n^3 + 1/2*n^2 + 1/6*n$.