Binomials and nCr notations (1 Viewer)

Narkis · Oct 8, 2017

I have no idea how to do them, please help me solve these questions.

1) nC1+nC2=3n
2)Find the constant term in the expansion of (x+2)⁶((1/x²)-1)²

And finally,
3) Given that (2+(x/10))ⁿ=1024+ax+bx²+terms involving higher powers of x.
(a) Find the value of n.
(b) Find the values of a and b.

These questions are actually really basic and I know that..........but I just keep getting the wrong answers, anyone who can solve them is literally my superstar. (orz)

Drongoski · Oct 8, 2017

$\binom n 1 + \binom n 2 = 3n \implies \frac {n}{1!} + \frac {n(n-1)}{2!} = 3n \implies \frac {n}{1} + \frac {n(n-1)}{2} = 3n \\ \\ \implies 2n + n^2 - n = 6n \implies n(n-5) = 0 \implies n = 5 $ or $ n = 0 $ (can ignore this case)$ \\ \\ 2) \\ $expression$ = \sum _ {i = 0} ^6 \binom 6 i x^{6-i} 2^i \times (x^{-4} -2x^{-2} + 1) \\ \\ $constant term $ = \binom 6 2 2^2x^4 \times x^{-4} + \binom 6 4 2^4 x^2 \times {(-2x^{-2})} + 2^6 \times 1 \\ \\ = 4 \binom 6 2 - 32 \binom 6 4 + 64 = -356$

3)

$\left ( 2 + 0.1x \right )^n = 2^n + \binom n 1 2^{n-1}(0.1x)^1 + \binom n 2 2^{n-2} (0.1x)^2 + \cdots \\ \\ \therefore 2^n = 1024 \implies n = 10 \implies ax = 10 \times 2^9 \times 0.1 x \implies a = 512\\ \\ \implies b = \binom {10} 2 \times 2^8 \times 0.1^2 = 115.2$

Binomials and nCr notations (1 Viewer)

Narkis

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Drongoski

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