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Question

The value of C20+3C21+5C22+ up to 51 terms is equal to
( where Cr= 50Cr)

A
51250
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B
102100C50
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C
51100C50
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D
99100C50
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Solution

The correct option is C 51100C50
Let S=C20+3C21+5C22++101C250
Now,
S=101C250+99C249+97C248++C20
Adding both, we get
2S=102[C20+C21+C22++C250]S=51 100C50

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