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 $$P^{2} - Q^{2} =$$

Solution

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

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

$$(a - x)^{n}$$   =   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 both we get 

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

hence  

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

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

there fore PQ we get 

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

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

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

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

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


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