Thread: Sigma notation & Pascal's Triangle/Binomial Theorem.

I have no idea how to do these:

A) "Find the sum of the arithmetic series"




B) "Expand (x+2)^4"

So far, all I could manage to get is 1x^4(2)^0+4/1x^3(3). Is there an easier way to do this question without it becoming so messy/confusing?

Thanks in advance!

Sn= (N/2)(A1 + An)

Originally Posted by Mulya66

B) "Expand (x+2)^4"

So far, all I could manage to get is 1x^4(2)^0+4/1x^3(3). Is there an easier way to do this question without it becoming so messy/confusing?

Define
[nCr] = n!/[(n - r)!r!]

Then
(x + a)^n = [nC0]*x^n*a^0 + [nC1]*x^{n - 1}*a^1 + [nC2]*x^{n - 2}*a^2 + ... + [nC(n-1)]*x^1*a^{n - 1} + [nCn]*x^0*a^n

In your case:
(x + 2)^4 = [4C0]x^4 + [4C1](2)x^3 + [4C2](2)^2x^2 + [4C3](2)^3x + [4C4](2)^4

Now
[4C0] = 1
[4C1] = 4
[4C2] = 6
[4C3] = 4
[4C4] = 1

So
(x + 2)^4 = x^4 + 4*2x^3 + 6*4x^2 + 4*8x + 16

= x^4 + 8x^3 + 24x^2 + 32x + 16

-Dan

Originally Posted by Logarithm Freak

Sn= (N/2)(A1 + An)

Here's what I got:

n = 48

A1 = 4(1)-1
A1 = 3

A48 = 4(48)-1
A48 = 191

So,
S = 48/2(3+191)
S = 24(194)
S = 4656

Please let me know if I got this right.

Originally Posted by Mulya66

A) "Find the sum of the arithmetic series"

The more traditional way:

Sum[4n - 1, 1, 48] = Sum[4n, 1, 48] - Sum[1, 1, 48]

= 4*Sum[n, 1, 48] - Sum[1, 1, 48]

Now,
Sum[n, 1, N] = (N/2)(N + 1)
Sum[1, 1, N] = N

So
Sum[4n - 1, 1, 48] = 4*Sum[n, 1, 48] - Sum[1, 1, 48]

= 4*(48/2)(48 + 1) - 48

= 4704 - 48

= 4656

Originally Posted by Logarithm Freak

Sn= (N/2)(A1 + An)

Don't just give an answer like this explain where it comes from, and for that
matter what the variables/symbols represent.

RonL