If in the expansion of $(1 + x)^{n},$ the coefficient of pth and qth terms are equal and the relation between p and q is p-q =n . Find the value of $p^{2} + q^{2}$ . where $p \neq q .$

Or

Find the term independent of x in the expansion of ${(2 x - \frac{1}{x})}^{10} .$

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Solution

We know that the cofficient of rth term in the expansion of $(1 + x)^{n} i s^{n} C_{r - 1} .$

Therefore, coefficients of pth and qth terms in the expansion of $(1 + x)^{n} a r e^{n} C_{p - 1} .$ , respectively.

It is given that these coefficients are equal.

$∴^{n} C_{p - 1} . =^{n} C_{q - 1}$

$\Rightarrow p - a + q - 1 = n [∵^{n} C_{x} =^{n} C_{y} \Rightarrow x + y = n]$

$\Rightarrow P + q = n + 2 (i)$

Also given, $\Rightarrow p - q = n (i i)$

On adding Eqs,(i) and (ii) we get

$2 p = 2 n + 2 \Rightarrow p = n + 1$

On putting p=n+1 in Eq.(i), we get

$q = n + 2 - (n + 1) = 1$

Now, $p^{2} + q^{2} = (N + 1)^{2} + 1^{2}$

$= n^{2} + 1 + 2 n + 1 = n^{2} + 2 n + 2$

Or

Let (r+1) th term be independent of x in the given expansion ${(2 x - \frac{1}{x})}^{10} .$

Then , $T_{r + 1} =^{10} C_{r} (2 x)^{10 - r} {(- \frac{1}{x})}^{r}$

$\Rightarrow T_{r + 1} = (- 1)^{r}^{10} C_{r} (2)^{10 - r} (X)^{10 - 2 r}$

For this term to be independent of x, we must have

$10 - 2 r = 0 \Rightarrow r = 5$

So, $(5 + 1)$ i.e. 6 th terms is independent of x, i.e $T_{6} = (- 1)^{5}^{10} C_{5} (2)^{10 - 5}$

$= - \frac{10!}{5! 5!} \times 2^{5} = - 252 \times 32 = - 8064$

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Similar questions

If in the expansion of $(1 + x)^{n}$ , the coefficients of pth and qth terms are equal, prove that p+q =n+2, where $p \neq q .$

{(1 + x)}^{n}

p^{t h}

(p + 1)^{t h}

p

q

p + q =

(1 + x)^{n}

p^{t h}

(p + 1)^{t h}

p

q

p + q =

{(1 + x)}^{n}

(p + 1)

p \neq q

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