Binomials: From [(1+x)^n]^2 = (1+x)^{2n}, prove (1+nC1+nC2+...+nCn)^2 = 1+2nC1+...

lisaaa

New member
I need help with 5 and 6 pleaseee

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tkhunny

Moderator
Staff member
Can you show the seven terms of 5a?

pka

Elite Member
I need help with 5 and 6
Here is #5 You can use it as a model.
$$\displaystyle {(1 + 2x)^6} = \sum\limits_{k = 0}^6 {\dbinom{6}{k}{{(1)}^k}{{(2x)}^{6 - k}}}$$ now if x=1 we get $$\displaystyle {(1 + 2)^6} = \sum\limits_{k = 0}^6 {\dbinom{6}{k}{{(1)}^k}{{(2)}^{6 - k}}}$$

lisaaa

New member
Here is #5 You can use it as a model.
$$\displaystyle {(1 + 2x)^6} = \sum\limits_{k = 0}^6 {\dbinom{6}{k}{{(1)}^k}{{(2x)}^{6 - k}}}$$ now if x=1 we get $$\displaystyle {(1 + 2)^6} = \sum\limits_{k = 0}^6 {\dbinom{6}{k}{{(1)}^k}{{(2)}^{6 - k}}}$$
Im sorry but is there any other way to do this as I have not learnt what ∑ stands for yet. Thanks

Harry_the_cat

Senior Member
Can you expand (a+b)^6 using the binomial expansion?

lisaaa

New member
Can you expand (a+b)^6 using the binomial expansion?
would this be the correct expansion?

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tkhunny

Moderator
Staff member
So close!! Try (2x)^0, (2x)^1, (2x)^2, etc. The parentheses make a BIG difference. Look at pka's demonstration. You will see them, there.

JeffM

Elite Member
Im sorry but is there any other way to do this as I have not learnt what ∑ stands for yet. Thanks
$$\displaystyle \Sigma$$ is the capital letter sigma in the Greek alphabet (the equivalent of S in the Roman alphabet) and means to SUM (add up) (1+n-m) terms by substituting first m for k, then m+1 for k, and so on until you substitute n for k. So

$$\displaystyle \sum_{k=1}^1 2^k = 2^1 = 2.$$

$$\displaystyle \sum_{k=1}^2 2^k = 2^1 + 2^2 = 2 + 4 = 6.$$

$$\displaystyle \sum_{k=0}^3 2^k = 2^0 + 2^1 + 2^2 + 2^3 = 1 + 2 + 4 + 8 = 15.$$

For a more complicated example that is pertinent to your question

$$\displaystyle \left ( \sum_{k=0}^4 \dbinom{4}{k} * (-\ 1)^k * a^{(4-k)} * b^k \right) \equiv$$

$$\displaystyle \left ( (-\ 1)^0 * \dbinom{4}{0} * a^{(4-0)} * b^0 \right ) + \left ( (-\ 1)^1 * \dbinom{4}{1} * a^{(4-1)} * b^1 \right ) + \left ( (-\ 1)^2 * \dbinom{4}{2} * a^{(4-2)} * b^2 \right ) +$$

$$\displaystyle \left ( (-\ 1)^3 * \dbinom{4}{3} * a^{(4-3)} * b^3 \right ) + \left ( (-\ 1)^4 * \dbinom{4}{4} * a^{(4-4)} * b^4 \right ) =$$

$$\displaystyle \dfrac{4! * a^4b^0}{0! * (4 - 0)!} -\dfrac{4! * a^3b^1}{1! * (4 - 1)!} + \dfrac{4! * a^2b^2}{2! * (4 - 2)!} - \dfrac{4! * a^1b^3}{3! * (4 - 3)!} + \dfrac{4! * a^1b^0}{4! * (4 - 0)!} =$$

$$\displaystyle a^4 - 4a^3b + 6a^2b^2 - 4ab^3 + b^4 = (a - b)^4.$$

pka

Elite Member
Im sorry but is there any other way to do this as I have not learnt what ∑ stands for yet. Thanks
Because I am curious and for no other reason, I have a question for lisaaa.
Given that you say that notation $$\displaystyle \sum\limits_{k =1 }^6 {{t_k}}$$ is not known to you, why post in the Advances Mathematics subforum?
Moreover, it is odd that you are asked to tackle this topic without being given the tools to understand the help received.