Find the general formula for the sequences

Question

1=1

2+3+4+=1+8

5+6+7+8+9=8+27

10+11+12+13+14+15+16=27+64

Find the formula is suggested by these equations?Prove your answer is correct.

I saw this question on practice exam and the answer is = (2n-1)(n²-n+1) but I only know a few steps and didn't get this answer. Please explain. Thanks

Answer 1

Do you know the formula for arithmetic sequence? $${(a_1+a_n)n\over2}$$

You need to apply it to find the first number: $(1+2n-3)(n-1)\over2$$=(n-1)^2+1$

And you need to apply it again to find the sum: $((n-1)^2+1+(n-1)^2+2n-1)(2n-1)\over2$$=(2n-1)(n^2-n+1)$

Answer 2

Each equation is the sum of the terms from the number after the last perfect square to the next perfect square. The sum of consecutive integers from $a_1$ to $a_m$ is given by $\frac{m}{2}(a_1+a_m)$ where m is the number of terms being added up.

Now if $n$ is the equation number, the first term, $a_1$, is one more than the previous square, $(n-1)^2+1$, and the last term, $a_m$, is the number squared ($n^2$). The number of terms is the odd sequence, so $m=2n-1$. Putting this together we have:

$$\frac{2n-1}{2}\left((n-1)^2+1+n^2\right)$$ $$=\left(2n-1\right)\left(n^2-n+1\right)$$

If we multiply and manipulate this it gives the pattern on the right:

$$2n^3-3n^2+3n-1$$ $$=n^3-3n^2+3n-1+n^3$$ $$=(n-1)^3+n^3$$

Answer 3

Let us denote $q$ the index of an equation (starting from $q=1$), and the LHS is

$$(p^2+1)+(p^2+2)+\cdots+q^2,$$ where $p=q-1$. Indeed, the sequence of all LHS terms is continuous and there are $q^2-p^2$ of them, which equals $q+p$ and increases by two every time.

As we have an arithmetic progression, by pairing opposite terms we see that we can replace all of them by the middle one, $\dfrac{(p^2+1)+q^2}2=q^2-q+1=q^2-pq+p^2$.

Then, the sum of the terms equals

$$(q+p)(q^2-pq+p^2)=p^3+q^3.$$

Answer 4

$$\color{blue}1=1\\ 2+\color{blue}3+4=3+3+3\\ 5+6+\color{blue}7+8+9=7+7+7+7+7\\ 10+11+12+\color{blue}{13}+14+15+16=13+13+13+13+13+13+13\\ 17+18+19+20+\color{blue}{21}+22+23+24+25=21+21+21+21+21+21+21+21+21$$

$$\begin{align} 1\cdot1&=(1+0)(1^2-0)=(1+0)(1^2-1\cdot0+0^2)=1^3+0^3\\ 3\cdot3&=(2+1)(2^2-1)=(2+1)(2^2-2\cdot1+1^2)=2^3+1^3\\ 5\cdot7&=(3+2)(3^2-2)=(3+2)(3^2-3\cdot2+2^2)=3^3+2^3\\ 7\cdot13&=(4+3)(4^2-3)=(4+3)(4^2-4\cdot3+3^2)=4^3+3^3\\ 9\cdot21&=(5+4)(5^2-4)=(5+4)(5^2-5\cdot4+4^2)=5^3+4^3\\ \end{align}$$