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If we tried to sum a series where the difference of the difference of the difference is constant, i.e. sum of pyramidal numbers, the result would be a pentatope number. And so on... An example of the summation of pyramidal numbers, extending from the original question, would be
$\begingroup$ Hint: reverse the series and sum it up term by term with the original series. So $\cos(a)+\cos(a+(n-1)\cdot d)$, etc... And use the Simpson formula for sums of cosines (and sines for the other identity). $\endgroup$ –
Finding the sum of arithmetic series when last term and common difference is given . 4.
Sum of Arithmetic series. Ask Question Asked 9 years, 7 months ago. Modified 9 years, 1 month ago. Viewed ...
How the heck do I find the sum of a series like $\sum\limits_{n=3}^\infty\frac{5}{36n^{2}-9}$? I can't seem to convert this to a geometric series and I don't have a finite number of partial sums, so I'm stumped.
For example: $$1+2+\\text{...}+n=\\frac{n(n+1)}{2}~~~(1)$$ $$1^2+2^2+\\text{...}+n^2=\\frac{n(n+1)(2n+1)}{6}~~~(2)$$ In this equality, I sometimes recall by heart ...
4th term of arithmetic progression is the sum of squares of the first three terms. Hot Network Questions What does the absence of a ground state physically mean?
Additional properties of finite series that have the property: $\sum a_k = -\sum \frac{a_k}{k}$ 2 How to find fourth term a of geometric series using sum of first three terms and second term?
There is an easier way. Notice that: $$\sum_{r = 1}^n a + (r-1)d = \sum_{r = 1}^n a + \sum_{r = 1}^n rd - \sum_{r = 1}^n d = na + d\left (\sum_{r = 1}^n r\right) - nd $$ Now, just prove by induction that $$\sum_{r = 1}^n r = \frac{n(n+1)}{2}$$ which is much easier, and manipulate the previous expression to get what you need.
Finding the Sum of an arithmetic series when the sum of the first two terms is given and the 20 th term is 93. 1 General formula for the power series of $\dfrac{1}{(1+x)^3}$