Summing $3+7+14+24+37...$ up to $10$ terms

Question

What is $3+7+14+24+37...$ up to $10$ terms?

I only see that difference between two consecutive terms is in AP ie $7-3=4,14-7=7,24-14=10$ and so on. Any ideas on how to do it?

Answer 1

The sequence and its forward differences:

$$a_n : 3,7,14,24,37,..$$

$$\Delta a_n : 4,7,10,13,..$$

$$\Delta^2 a_n : 3,3,3,...$$

$$\Delta^3 a_n : 0,0,..$$

$$\Delta^4 a_n : 0,..$$

Now we make a reasonable guess that $\Delta^4 a_n=0$ i.e. we guess $\Delta^4 a_n$ is a sequence with all terms as $0$ to get $\Delta^k a_n=0$ when $k \geq 4$ and is an integer:

$$\Delta^5 a_n=0 : 0,..$$

$$\Delta^6 a_n=0 : 0,...$$ ..

If we take $a_0$ to be the first term of the sequence than looking at the first term of each of the above sequences $3,4,3,0,0,0,..$ we have:

$$a_n=3{n \choose 0}+4{n \choose 1}+3{n \choose 2}+0{n \choose 3}+0{n \choose 4}+0{n \choose 5}+\cdots$$

$$=3{n \choose 0}+ 4{n \choose 1}+3{n \choose 2}$$

Where now you can apply the summation formula $\sum_{n=0}^{x} {n \choose k}={x+1 \choose k+1}$. You can shift this to the right one if you want $a_1$ to represent the first term instead of $a_0$. Either way, you will get that the sum you want is equal to :

$$3{10 \choose 1}+4{10 \choose 2}+3{10 \choose 3}=570$$

Because there are $10$ terms $a_0,..a_9$ and using the given formula that is what we get.

If you want a proof of what I hope you are observing with the differences let me know. The summation formula was grabbed from Wikipedia.

Answer 2

Since the second differences are constant, the terms come from a quadratic polynomial. One can check that the terms of the series come from the polynomial $1.5k^2-0.5k+2$ where $k\geq 1$. Then $$\sum\limits_{k=1}^{10} (1.5k^2-0.5k+2)=1.5\sum\limits_{k=1}^{10} k^2-0.5\sum\limits_{k=1}^{10} k+2\sum\limits_{k=1}^{10}1$$ which you can evaluate using the well-known sum formulas.

Answer 3

$x=1 \phantom{n} 2 \phantom{n} 3 \phantom{n} 4$

$y=3 \phantom{n} 7 \phantom{n} 14 \phantom{n} 24$

$y=m_2(x-x_1)(x-x_2)+m_1(x-x_1)+y_1$

$m_1$ is slope

$m_1=(y_2-y_1)/(x_2-x_1)$

$m_1=(7-3)/(2-1)$

$m_1=4$

$m_2$ changes in slope

$m_2=((y_3-y_2)-(y_2-y_1))/(x_3-x_1)$

$m_2=((14-7)-(7-3))/(3-1)$

$m_2=3/2$

$y=3/2*(x-2)(x-1)+2*(x-1)+3$

$y=1.5x^2-0.5x+4$

Answer 4

Take a difference of the difference, and you get a constant sequence. This is a quadratic polynomial, not a linear one...

Answer 5

This answer seeks to expand on the "one can check that" comment in Foobaz's answer. First, let $A=\{3,7,14,24,37,\ldots\}$. The pattern for the difference, $d_n$, between terms $n$ and $n+1$ is clear enough: $$ d_n=4+(n-1)3=1+3n. $$ Hence, $$ a_{n+1}=\frac{n}{2}(5+3n)+3 $$ or, more usefully, \begin{align} a_n&=\frac{n-1}{2}(2+3n)+3\\[1em] &=(n-1)+\frac{3n^2-3n}{2}+3\\[1em] &= \frac{2n-2+3n^2-3n+6}{2}\\[1em] &= \frac{3n^2-n+4}{2}. \end{align} Thus, we have the following: \begin{align} S_{10}&= \sum_{i=1}^{10}\frac{3i^2-i+4}{2}\\[1em] &= \frac{3}{2}\sum_{i=1}^{10}i^2-\frac{1}{2}\sum_{i=1}^{10}i+2\sum_{i=1}^{10}1\\[1em] &= \frac{3}{2}\left[\frac{10(10+1)(2(10)+1)}{6}\right]-\frac{1}{2}\left[\frac{10(10+1)}{2}\right]+2(10)\\[1em] &= 570. \end{align}