Formation of a Differential Equation from a General Solution
If 1+xn=C0+...
Question
If (1+x)n=C0+C1x+C2x2+....+Cnxn, then the value of C0+2C1+3C2+....+(n+1)Cn will be
A
(n+2)2n−1
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B
(n+1)2n
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C
(n+1)2n−1
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D
(n+2)2n
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Solution
The correct option is A(n+2)2n−1 Since we know that 2n=Cn0+Cn1+Cn2+Cn3+Cn4+Cn5+....=∑r=nr=oCnrCn0+2Cn1+3Cn2+4Cn3+5Cn4+6Cn5+....=∑nr=0(r+1)Cnr=∑nr=0(r)Cnr+∑r=nr=oCnr⟹rCnr=nCn−1r−1∴∑nr=0(r+1)Cnr=n2n−1+2n=(n+2)2n−1