Question 131

If P is the sum of odd terms and Q is the sum of even terms in the expansion of $$(x + y)^n$$ then $$P^{2} - Q^{2} =$$

we know that the expansion of sum of even term and sum of odd term are respective given belowΒ 

$$(a + x)^{n}$$Β  =Β  n$$C_{0}$$ $$a^{n}$$ +Β  n$$C_{1}$$ $$a^{n-1}$$x +Β  n$$C_{2}$$ $$a^{n-2}$$ $$x^{2}$$Β  + ............. + n$$C_{n}$$ $$x^{n}$$

$$(a - x)^{n}$$Β Β  =Β  Β n$$C_{0}$$ $$a^{n}$$ - n$$C_{1}$$ $$a^{n-1}$$x + n$$C_{2}$$ $$a^{n-2}$$ $$x^{2}$$ - ............. + $$(-1)^n$$Β n$$C_{n}$$ $$x^{n}$$

adding both we getΒ 

$$(a + x)^{n}$$ +Β $$(a - x)^{n}$$ = 2(Β n$$C_{0}$$ $$a^{n}$$ +Β  n$$C_{2}$$ $$a^{n-2}$$ $$x^{2}$$ + ............. +Β  n$$C_{n}$$ $$x^{n}$$)

henceΒ Β 

P = ($$(a + x)^{n}$$ +Β $$(a - x)^{n}$$)\2Β  Β  Β  Β  equation 1Β Β 

QΒ = ($$(a + x)^{n}$$ - $$(a - x)^{n}$$)\2Β  Β  Β  Β Β equation 2Β 

there fore PQ we getΒ 

PQ =Β Β  ($$(a + x)^{n}$$ + $$(a - x)^{n}$$)\2 $$\times$$Β ($$(a + x)^{n}$$ - $$(a - x)^{n}$$)\2Β  =Β ($$(a + x)^{2n}$$ + $$(a - x)^{2n}$$)\4

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

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

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

$$P^{2}$$ - $$Q^{2}$$ =Β Β ($$(a)^{2}$$ -Β $$(x)^{2}$$)^nΒ Β Answer

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