Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

We have the succession and its formula: $$ 1^2+4^2+\cdots+ (3k-2)^2 = \dfrac{k(6k^2-3k-1)}{2} $$

Now we need to apply it for $k+1$: $$ 1^2+4^2+\cdots+ (3n-2)^2 +(3(k+1)-2)^2 = \\ \dfrac{k(6k^2-3k-1)}{2} + (3(k+1)-2)^2 $$

I know that the result must be $\frac{1}{2}(k+1)(6(k+1)^2-3(k+1)-1)$ but I wasn't able to find the elegant solution during the test. I can't even see from where to factor out $k+1$.

Care to give a hint?

share|improve this question
3  
The last term is $(3k+1)^2$. Expand. Bring whole expression to common denominator $2$. Simplify the top. Since you expect a $k+1$ term, divide the cubic on top by $k+1$ (polynomial division). Miraculously it divides exactly. –  André Nicolas Nov 28 '12 at 21:18
add comment

2 Answers

up vote 2 down vote accepted

First combine everything into a single fraction:

$$\begin{align*} \frac{k(6k^2-3k-1)}{2} + \big(3(k+1)-2\big)^2&=\frac{k(6k^2-3k-1)}2+(3k+1)^2\\ &=\frac{k(6k^2-3k-1)+2(3k+1)^2}2\\ &=\frac{6k^3-3k^2-k+18k^2+12k+2}2\\ &=\frac{6k^3+15k^2+11k+2}2\;.\tag{1} \end{align*}$$

At this point the easiest thing to do is to multiply out the desired expression,

$$\frac{(k+1)\left(6(k+1)^2-3(k+1)-1\right)}2\;,$$

and verify that it’s equal to $(1)$. If you want to keep working forward from $(1)$, however, you can. We know that $k+1$ should be a factor of the numerator, so we could simply try to divide it out, but we can easily confirm this: by inspection $k=-1$ is a zero of the numerator, so $k-(-1)=k+1$ is a factor. Ordinary polynomial long division or synthetic division quickly yields the factorization

$$6k^3+15k^2+11k+2=(k+1)(6k^2+9k+2)\;.$$

$6(k+1)^2=6k^2+12k+6$, so $6k^2+9k+2=6(k+1)^2-3k-4=6(k+1)^2-3(k+1)-1$, as desired.

share|improve this answer
add comment

Hint $\ $ Your question amounts to verifying the following polynomial equality

$$\rm (k\!+\!1)\, f(k\!+\!1) - k\, f(k) =\, 2\,(3k\!+\!1)^2\quad for\quad f(k) =\, 6k^2\! - 3k - 1$$

Since LHS and RHS are polynomials of degree $\color{#C00} 2,$ to prove that they are equal it suffices to check that they have equal values at $\,\color{#C00}3\,$ points. That's easy: $ $ LHS = RHS $ $ holds for $\rm\:k = -1,0,1,\:$ viz. $$\begin{eqnarray}&&\rm k=-1:\ \ f(-1) = 8 \\ &&\rm k\ =\ 0:\ \ \ f(1) = 2\\ &&\rm k\ =\ 1:\ \ \ 2\, f(2)\!-\!f(1) = 32 \end{eqnarray}$$

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.