No links please Q If 1 + xn = C0 + C1x + C2x2 + - Maths - Relations and Functions

Question

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Q     I f   1   +   x n   =   C
0   +   C 1 x   +   C 2 x 2   +  
  +   C n x n ,   p r o v e   t h a t , ( i )
    C 0   +   2 C 1   +   3   C
2   +     +   n   +   1   C n
  =   n   +   2   ·   2 n - 1

Muskaan Talwar · Accepted Answer

Dear Student,

Given that:1+xn=C0+C1x+C2x2+
+CnxnMultipying both sides with x, we getx1+xn=C0x+C1x2+C2x3+
+Cnxn+1Differentiating both sides w r t
 x, we get1+xn+nx1+xn-1=C0+2C1x+3C2x2+
+n+1CnxnPut x=1 in above equation, we get1+1n+n1+1n-1=C0+2C1+3C2+
+n+1Cn⇒2n+n2n-1=C0+2C1+3C2+ +n+1Cn⇒2n-12+n=C0+2C1+3C2+
+n+1CnThus, C0+2C1+3C2+ +n+1Cn=n+22n-1

Regards

No links please Q . I f 1 + x n = C 0 + C 1 x + C 2 x 2 + . . . + C n x n , p r o v e t h a t , ( i ) C 0 + 2 C 1 + 3 C 2 + . . . + n + 1 C n = n + 2 · 2 n - 1

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$Q . I f {(1 + x)}^{n} = C_{0} + C_{1} x + C_{2} x^{2} + . . . + C_{n} x^{n}, p r o v e t h a t, (i) C_{0} + 2 C_{1} + 3 C_{2} + . . . + (n + 1) C_{n} = (n + 2) \cdot 2^{n - 1}$