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Question

The sum mi=0(10i)(20mi), (where (pq=0 if p<q)), is maximum when m is

A
5
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B
15
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C
10
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D
20
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Solution

The correct option is B 15
For m=5, 5r=0(10i)(205i)
=(100)(205)+(101)(204)++(105)(200)
for m=10, 5i=0(10i)(2010i)
=(100)(2010)+(101)(209)+(102)(208)++(1010)(200)
for m=15,15i=0(10i)(2015i)
=(100)(2015)+(101)(2014)+(102)(2013)++(1010)(205)
and for m = 20, 20i=0(10i)(2020i)
=(100)(2020)+(101)(2019)++(1010)(2010)
Clearly, the sum is maximum for m = 15
Note that 10Cr is maximum for r = 5 and 20Cr is maximum for r = 10. Note that the single term 10C5×20C10 (in case m = 15) is greater than the sum
10C200C10+10C201C9+10C202C8+
10C208C2+10C209C1+10C2010C0 (in case m = 10).
Also the sum in case m = 10 is same as that in case m = 20

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