22.C01.2+23.C12.3+....2n+2Cn(n+1)(n+2)

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3n+2−2n−5(n+1)(n+2)

3n+2+2n+5(n+1)(n+2)

3n+5−5n+3(n+1)2

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answer is A.

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Detailed Solution

We have 1+xn=C0+C1x+C2x2+.......+Cnxn Integrating w.r.t x we get, 1+xn+1n+1=C0x+C1x22+C2x33+.......+Cnxn+1n+1+C where C is an arbitrary constant Putting x=0, we get C=1n+1 ∴1+xn+1n+1=C0x+C1x22+C2x33 ......... +Cnxn+1n+1+1n+1 Again, integrating w.r.t x we get 1+xn+2n+1n+2=C0x22+C1x32.3+C2x43.4+.......+Cnxn+2n+1n+2+1n+1x+k where k is an arbitrary constant. For x=0, we get k=1n+1n+2 ∴1+xn+2n+1n+2=C0x22+C1x32.3+C2x43.4+.......+Cnxn+2n+1n+2+1n+1x+1n+1n+2 Now for x=2, we have C0222+C1232.3+C2243.4+.......+Cn2n+2n+1n+2= 3n+2-2n-3n+1n+2

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