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Question

State whether the following statement is true or false.
(1+x)n=P0+P1x+P2x2+Pnxn, then:
P0P2+P4+....=2n2cosnπ4

A
True
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B
False
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Solution

The correct option is A True
Given,
(1+x)n=P0+P1x+P2x2+Pnxn.............[1]

Putting x=i in eq.[1] :-

(1+i)n=P0+P1(i)+P2(i)2..........+Pn(i)n

(1+i)n=P0+iP1P2iP3+P4..........Pn(i)n..............................[2]

Now putting x=i in eq.[1] :-

(1i)n=P0+P1(i)P2+iP3+P4...............Pn(i)n

(1i)n=P0P1+P2.............+(1)nPnxn.............................[3]

Adding equations [2] and [3]:

(1+i)n+(1i)n=2[P0P2+P4.........]

[2ei(π/4)]n+[2ei(π/4)]n=2[P0P2+P4.........]

(2)n2ei(nπ/4)+(2)n2ei(nπ/4)=2[P0P2+P4.........]

(2)n2[cos n(π/4)+isin n(π/4)+cos n(π/4)sin n(π/4)]=2[P0P2+P4.........]

(2)n2[2cos nπ4]=2[P0P2+P4........]

Hence,

(2)n2[cos nπ4]=[P0P2+P4........]

So, above statement is true.

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