Telescoping Series

Find $S_r = 1^r + 2^r + 3^r + ... + n^r$ where r is any fixed positive integer in terms of $S_1, S_2, ... S_{r-1}$.

Degree Ceremony

What does Pythagoras' Theorem tell you about these angles: 90°, (45+x)° and (45-x)° in a triangle?

OK! Now Prove It

Make a conjecture about the sum of the squares of the odd positive integers. Can you prove it?

Sum the Series

Stage: 5

Published May 1998,February 2011.

The Sum of 1 + 22 + 333 + 4444 + ... to $n$ terms

This is an extension of the NRICH problem called Clickety Click and All the Sixes , where the solution involves summing the series
$$1 + 11 + 111 + 1111 +\cdots$$ to $n$ terms.

Each term of the series $1 + 22 + 333 + 4444 +\cdots$ can be written as $k (1 + 10 + 100 + \cdots + 10^{k -1} )$ for some value of $k$. Using geometric series theory each term can be written as $k (10^ {k} -1)/9$. So the sum $S_n$ can be written as \eqalign{ S_n &=& 1\left(\frac{10^1-1}{9}\right) + 2\left(\frac{10^2-1}{9}\right) + 3\left(\frac{10^3-1}{9}\right) + 4\left(\frac{10^4-1}{9}\right) + \cdots \\ \; &=& {1\over9}(1.10^1 + 2.10^2 + 3.10^3 +4.10^4 + \cdots + n.10^n - (1+2+3+4+ \cdots +n)) \\ \; &=& {1\over9}(1.10^1 + 2.10^2 + 3.10^3 +4.10^4 + \cdots + n.10^n - {1\over2}n(n+1))}

To find the formula for $1.10^1 + 2.10^2 + 3.10^3 + 4.10^4 + \cdots + nx^n$ consider the series $$y = 1+ x + x^2 + x^3 + \cdots + x^n = \frac{x^{n+1}-1}{x-1}$$

Differentiating this expression and multiplying the derivative by x we get \eqalign{ x\frac{dy}{dx} &=& x + 2x^2 + 3x^3 + \cdots + nx^n \\ \; &=& \frac{nx^{n+2} - (n+1)x^{n+1} + x}{(x-1)^2}}.

Hence $$10 + 2.10^2 + 3.10^3 + \cdots + n.10^n = \frac{n.10^{n+2} - (n+1).10^{n+1} + 10}{81}$$

So the sum of $n$ terms of the original series is $$S_n = 1 + 22 + 333 + 4444 + \cdots = {1\over9}\left(\frac{n.10^{n+2} - (n+1).10^{n+1} + 10}{81} - {1\over2}n(n+1)\right)$$

Note that when $n$ is greater than 9, terms cannot be written with a repeated single digit and the $k^{th}$ term should be treated as $k (10^0 + 10^1 + 10^2 + \cdots + 10^{k-1})$.