Binomial Theorem Questions (1 Viewer)

Bored Of Fail · Feb 15, 2011

binomials are so easy, what is hard bout them

Trebla · Feb 15, 2011

UnrealAnchovies said:
Hiya,

I have a few BT questions which I'm not able to complete. They're probably pretty straightforward.

1) In the expansion of (1+x)^20 in ascending order of x show that there are two successive terms with the same coefficient, which is the greatest of any of the coefficients.

2) Prove that in the expansion of (1+2x)^9, when x = 1/3, there are two greatest terms.

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3) By equating the coefficient of x^r on each side of the identities:
(1+x)^(n+1) = (1+x)(1+x)^n and (1+x)^(n+2) = (1+x)^2(1+x)^n
show that n+1Cr = nCr + nCr-1 and n+2Cr = nCr + 2(nCr-1) + nCr-2

Thanks in advance.

Question 1

$T_k=\binom{20}{k-1}x^{k-1}\\\\T_{k+1}=\binom{20}{k}x^k\\\\$If there are two successive terms with the same coefficient then there exists a $k$ such that $\binom{20}{k-1}=\binom{20}{k}\\\\\Rightarrow \dfrac{20!}{(k-1)!(21-k)!}=\dfrac{20!}{k!(20-k)!}\\\\\Rightarrow 21-k=k\\\\\Rightarrow k = 10.5 \\\\$Since $k$ must be an integer there are no successive terms with equal coefficients$$

Question 2

Not sure about the explanation.

$T_k=\binom{9}{k-1}\left(\dfrac{2}{3}\right)^{k-1}\\\\T_{k+1}=\binom{9}{k}\left(\dfrac{2}{3}\right)^k\\\\$Consider the inequality $ T_{k+1}\geqT_k \\\\\Rightarrow \dfrac{9!}{k!(9-k)!}\times\dfrac{2}{3}\geq\dfrac{9!}{(k-1)!(10-k)!}\\\\\Rightarrow \dfrac{2}{3k}\geq \dfrac{1}{10-k}\\\\\Rightarrow 20-2k \geq 3k\\\\\Rightarrow k\leq 4\\\\$Greatest term occurs when $k=4$ which gives $\binom{9}{4}\left(\dfrac{2}{3}\right)^4\\\\$However, we could have easily used the strict inequality $T_{k+1}> T_k$ which leads to $k=3$ for the chosen greatest term which gives $\binom{9}{3}\left(\dfrac{2}{3}\right)^3\\\\$Both give the same answer.$$

Question 3

$(1+x)^{n+1}=(1+x)^n+x(1+x)^n\\\\\Rightarrow \displaystyle\sum_{k=0}^{n+1}\binom{n+1}{k}x^k=\displaystyle\sum_{i=0}^{n}\binom{n}{i}x^i+x\displaystyle\sum_{j=0}^{n}\binom{n}{j}x^j\\\\\Rightarrow \displaystyle\sum_{k=0}^{n+1}\binom{n+1}{k}x^k=\displaystyle\sum_{i=0}^{n}\binom{n}{i}x^i+\displaystyle\sum_{j=0}^{n}\binom{n}{j}x^{j+1}\\\\$Coefficient of $x^r$ on LHS is clearly $\binom{n+1}{r}\\\\$Coefficient of $x^r$ in $\displaystyle\sum_{i=0}^{n}\binom{n}{i}x^i$ occurs when $i=r$ hence coefficient is $\binom{n}{r}\\\\$Coefficient of $x^r$ in $\displaystyle\sum_{j=0}^{n}\binom{n}{j}x^{j+1} $ occurs when $j+1=r\Rightarrow j = r-1\\\\$Hence coefficient is $\binom{n}{r-1}\\\\$Coefficient of $\displaystyle\sum_{i=0}^{n}\binom{n}{i}x^i+\displaystyle\sum_{j=0}^{n}\binom{n}{j}x^{j+1}$ is therefore $\binom{n}{r}+\binom{n}{r-1}\\\\$Equating the coefficients leads to the result.$$

For the second part, it is pretty much the same approach noting that

$(1+x)^{n+2}=(1+x)^n+2x(1+x)^n+x^2(1+x)^n\\\\$

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Binomial Theorem Questions (1 Viewer)

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Trebla

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