La pendiente de la siguiente función lineal es:
f(x)=−32x+2
2 X 2
$$\sum_{k=1}^{n=1}{(-1)^{k-1}k^2}=(-1)^{n-1}[n(n+1)/2]$$
i have five digits the ten thousand digit is 900 divided by 100
The tens digit is one fifth of 25
The thousand digit is 2 lss t...
$$\sum_{k=1}^{n=1}{(-1)^{k-1}k^2=(-1)^{n-1}[n(n+1)/2]}$$
0.05x = 0.25y-1.80
0.05x = 0.25y+1.80
what is 1+1
$$\sum_{k=1}^{n=1}{log(k+1/k)=log(n+1)}$$