Write the following series in sigma notation: $3 + 10 + 21 + 36 + 55$

Question

What I need is an explanation of the method. I tried to get the general formula (for the nth term) of the series but I can't find it.

Answer 1

I hope you are able to see the pattern:

$3=1\cdot 3\\ 10 = 2 \cdot 5\\ 21 = 3 \cdot 7\\ 36 = 4 \cdot 9 \\ 55 = 5 \cdot 11$

Answer 2

Of course, given any finite number of values, there exist infinitely many functions that give those values. But, given $n$ points, there exists a unique polynomial of degree $n-1$ that gives those values. Here we are given five values, $3, 10, 21, 36$, and $55$ so, taking these as the points $(0, 3), (1, 10), (2, 21), (3, 36)$, and $(4, 55)$, there exists a polynomial of degree $4$ that gives those values. You could use "Newton's Divided Differences", as suggested by E.H.E. but a less sophisticated method is to write that polynomial as $ax^4+ bx^3+ cx^2+ dx+ e$ and use those values to get $5$ equations to solve for $a, b, c, d$, and $e$.

Taking $x= 0, e= 3$.

Taking $x= 1, a+ b+ c+ d+ e= 10$.

Taking $x= 2, 16a+ 8b+ 4c+ 2d+ e= 21$.

Taking $x= 3, 81a+ 27b+ 9c+ 3d+ e= 36$.

Taking $x= 4, 256a+ 64b+ 16c+ 4d+ e= 55$.

Solve those equations for $a, b, c, d$, and $e$.

Answer 3

Answer :

We can see that :

$3 = 1(1×2+1)$

$10 = 2(2×2+1)$

$21 = 3(2×3+1)$

$36 = 4(2×4+1)$

$55 = 5(2×5+1)$

After that we can see this numbers given by polynomial
$n(2×n+1)$ or $ 2n^2 +n$

Finally:

$\sum_{1}^{5} 2n^2 +n$