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binomial help again sorryyyy. (1 Viewer)
- Thread starter yashbb
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Life'sHard
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Let x = 1 on LHS and RHS
wait sorry, i dont get what u do after that
ooohhhhhh ok ok tysm!!!!!You basically just sub in
^n\\=2^n\\\\\text{RHS}=\binom{n}{0}+\binom{n}{1}(1)+\binom{n}{2}(1)^2+\binom{n}{3}(1)^3\text{...}+\binom{n}{n}(1)^n\\\\=\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\binom{n}{3}\text{...}+\binom{n}{n}\\\\=\sum_{k=0}^{n} \binom{n}{k}\\\\\\\therefore 2^n=\sum_{k=0}^{n} \binom{n}{k})
With these types of binomial identity questions, seeking that you find a relationship between binomial coefficients, there are essentially four approaches that you need to recognise:
} \qquad \text{Using sigma notation, state the binomial expansion of }\ (1+x)^n)
} \qquad \text{Show that }\ \sum_{k=0}^n \binom{n}{k} = 2^n)
} \qquad \text{Show that }\ \binom{n}{0} + \binom{n}{2} + \binom{n}{4} + ... = \binom{n}{1} + \binom{n}{3} + \binom{n}{3} + ...)
} \qquad \text{Find the value of }\ \sum_{k=0}^n \binom{2n}{2k})
Example Question 2
^n\ \text{as a series and hence show that }\ \sum_{k=1}^n k\binom{n}{k} = n2^{n-1})
- Substitute a value, as in the one above. This can be extended by combining results - example 1 below
- Use algebra or calculus on an expression prior to applying approach 1 - like example 2
- Form an expression by selecting terms from an identity and equating coefficients - like from the start of the 2020 MX1 HSC question
- Deriving results using combinatorics rather than, or accompanied by, algebra
Example Question 2