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Question

Let (1+2x)20=a0+a1x+a2x2+....+a20x20 Then, 3a0+2a1+3a2+2a3+3a4+2a5+...2a19+3a20 equals to:

A
5.32032
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B
5.320+32
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C
5.320+12
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D
5.32012
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Solution

The correct option is C 5.320+12
Given : (1+2x)20=a0+a1x+a2x2+......a20x20
(12x)20=a0a1x+a2x2+......a20x20
(1+2x)20+(12x)20
=2(a0+a2x2+a4x4......a20x20)
Take x=1
320+(1)20=2(a0+a2......a20)
32×{320+1}=32×2{a0+a2......a20}
32{320+1}=3{a0+a2......a20}
(1+2x)20(12x)20
=2(a1x+a3x3+......a19x19)
Take x=1
(1+2x)20(12x)20=2(a1+a3+a5......a19)
2(a1+a3+......a19)=320120=3201
To Find 3a0+2a1+3a2+2a3+.......
=3(a0+a2+......a20)+2(a1+a3+......a19)
=32{320+1}+3201
=320×52+12
Hence, the answer is 5.320+12.

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