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

tkhunny

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Can you show the seven terms of 5a?
 

pka

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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

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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

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

tkhunny

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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

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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

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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.
 
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