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Question

If (1+x)15=C0+C1x+C2x2+.....C15x15,
find the value of C2+2C3+3C4+.....14C15

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Solution

(1+x)15=C0+C1x+C2x2+....C15x15
(1+x)15x=C0x+C1+C2x+C3x2....C15x14
Differentiate both sides w.r.t. x
x15(1+x)141(1+x)15x2
=C0x2+C2+2C3x+3C4x2....14C15x13
Putting x = 1 on both sides
15.214215=C0+C2+2C3+3C4....14C15
214(152)+1=C2+2C3+3C4....14C15
The given series = 214 .13 + 1 = 219923.

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