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Sum of Coefficients of All Terms

In the expans...

Question

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

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

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

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

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

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Solution

The correct option is A $P^{2} - Q^{2} = {(x^{2} - a^{2})}^{n}$
$(x + a)^{n} =^{n} C_{0} x^{n} +^{n} C_{1} x^{n - 1} a^{1} +^{n} C_{2} x^{n - 2} a^{2} . . .^{n} C_{n} a^{n}$
$(x - a)^{n} =^{n} C_{0} x^{n} -^{n} C_{1} x^{n - 1} a^{1} +^{n} C_{2} x^{n - 2} a^{2} . . . (- 1)^{n}^{n} C_{n} a^{n}$
Therefore
$P = \frac{(x + a)^{n} + (x - a)^{n}}{2}$ ...(i)
$Q = \frac{(x + a)^{n} - (x - a)^{n}}{2}$ ...(ii)
Hence $4 (P^{2} - Q^{2}) = 4 (P + Q) (P - Q)$
$= 4 (x^{2} - a^{2})^{n}$
Therefore
$P^{2} - Q^{2} = (x^{2} - a^{2})^{n}$

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Similar questions

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:

P

Q

(x + a)^{n}

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

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

(x + a)^{n}

(P^{2} - Q^{2})

{(x + a)}^{n}

p^{2} + q^{2}

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