If $P$ and $Q$ are the sum of odd terms and the sum of even terms respectively, in the expansion of $(x + a)^{n}$ then prove that
(i) $P^{2} - Q^{2} = (x^{2} - a^{2})^{n}$
(ii) $4 P Q = (x + a)^{2 n} - (x - a)^{2 n}$

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Solution

$(a + x)^{n} =^{n} C_{0} a^{n} +^{n} C_{1} a^{n - 1} x +^{n} C_{2} a^{n - 2} x^{2} + . . .^{n} C_{n} x^{n}$
And
$(a - x)^{n} =^{n} C_{0} a^{n} -^{n} C_{1} a^{n - 1} x +^{n} C_{2} a^{n - 2} x^{2} + . . . (- 1)^{n}^{n} C_{n} x^{n}$
Adding, we get
$(a + x)^{n} + (a - x)^{n} = 2 [^{n} C_{0} a^{n} +^{n} C_{2} a^{n - 2} x^{2} +^{n} C_{4} a^{n - 4} x^{4} +^{n} C_{6} a^{n - 6} x^{6} + . . .^{n} C_{n} x^{n}]$
Hence
$P = \frac{(a + x)^{n} + (a - x)^{n}}{2}$ ...(i)
Similarly $Q = \frac{(a + x)^{n} - (a - x)^{n}}{2}$ ...(ii)
Therefore
$P Q = \frac{(a + x)^{n} + (a - x)^{n}}{2} \times \frac{(a + x)^{n} - (a - x)^{n}}{2}$
$= \frac{(a + x)^{2 n} - (a - x)^{2 n}}{4}$
Or
$4 P Q = (a + x)^{2 n} - (a - x)^{2 n}$
And
$P^{2} - Q^{2}$
$= (P + Q) (P - Q)$
$= [\frac{(a + x)^{n} + (a - x)^{n}}{2} + \frac{(a + x)^{n} - (a - x)^{n}}{2}] [\frac{(a + x)^{n} + (a - x)^{n}}{2} - \frac{(a + x)^{n} - (a - x)^{n}}{2}]$
$= (a + x)^{n} . (a - x)^{n}$
$= (a^{2} - x^{2})^{n}$
Hence proved.

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Similar questions

{(x + a)}^{n}

p^{2} + q^{2}

P

Q

(x + a)^{n},

P^{2} - Q^{2} = (x^{2} + a^{2})^{n}

In the expansion of $(x + a)^{n}$ , if the sum of odd terms be P and sum of even terms be Q. Consider the following statements:

(i) $P^{2} - Q^{2} = (x^{2} - a^{2})^{n}$

(ii) $P^{2} + Q^{2} = (x^{2} + a^{2})^{n}$

(iii) 4PQ = $(x + a)^{2 n} - (x - a)^{2 n}$

(iv) P - Q = $(x - a)^{n}$

Then the correct statement are:

(x + a)^{n}

(x + a)^{2 n} - (x - a)^{2 n}

P

Q

(x + a)^{n}

[(x + a)^{2 n} - (x - a)^{2 n}]

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