New answers tagged

1

$N-1\;\; x's$ of length $2 [1-2\;\;thru\;\; (N-1)-N]$ $N-2\;\; x's$ of length $3 [1-2\;\;thru\;\; (N-2)-N]$ ..... ..... $ 2\;\; x's$ of length $N-1 [1-(N-1)\;\;thru\;\; (N-1)-N]$ sum $ = [1+2+3+........(N-1)] - 1$ You must be knowing the formula for the sum of terms within the brackets $[\;\;\;]$ Added: a combinatorial approach Consider a string of ...


1

For $N$ x in a row, we have $N$ x, we have $N-1$ of 2 x's in a row, N-2 of 3 x's in a row. In general we have $N-k+1$ of $k$ x's in a row. To see why is that so, construct a window of length $k$ and move it.


0

Hint Since the terms have alternating sign you essentially want a simple form for expressions of the form $\sum_{n=0}^m ( (4n+1)^2 - (4n+3)^2 )$ and since $(4n+1)^2 - (4n+3)^2 = -4(4n+1)-4$ you only need to know $\sum_{n=0}^m n$ to reduce your sum to a simple form.


0

As we know, the $n^{th}$ partial sum for a given series is the sum of the first $n$ terms of the series. In our case the $n^{th}$ partial sum becomes $$S_{n}=1-2+2^2-2^3+\cdots+(-1)^{n-1} 2^{n-1}. $$ Multiplying the above equation by 2 we get $$2S_{n}=2-2^2+2^3-2^4+\cdots+(-1)^{n-1} 2^{n}.$$ Adding the above two equations simultinously, we obtain ...



Top 50 recent answers are included