The sum of the series 2⋅20C0+5⋅20C1+8⋅20C2+11⋅20C3+⋯+62⋅20C20
is equal to:
A
224
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B
225
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C
223
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D
226
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Solution
The correct option is B225 2⋅20C0+5⋅20C1+8⋅20C2+11⋅20C3+⋯+62⋅20C20 =(3⋅0+2)⋅20C0+(3⋅1+2)⋅20C1+(3⋅2+2)⋅20C2+1(3⋅3+2)⋅20C3+⋯+(3⋅20+2)⋅20C20=20∑r=0(3r+2)20Cr=320∑r=0r⋅20Cr+2⋅220 =320∑r=0r(20r)19Cr−1+221=60⋅219+221=225