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If f(m)=∑i=0m 3030−i20m−i where pq=pCq, then

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a

maximum value of f(m) is 50C25

b

f(0)+f(1)+⋯+f(50)=250

c

f(m) is always divisible by 50 (1≤m≤49)

d

The value of ∑m=050 (f(m))2=100C50

answer is A.

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Detailed Solution

f(m)=∑i=0m 3030−i20m−i=∑i=0m 30i20m−i=50Cmf(m) is greatest when m = 25. Also,f(0)+f(1)+⋯+f(50)=50C0+50C1+50C2+⋯+50C50=250Also, 50Cm s not divisible by 50 for all m as 50 is not a prime number∑m=050 (f(m))2= 50C02+ 50C12+ 50C22+⋯+ 50C502=100C50

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If f(m)=∑i=0m 3030−i20m−i where pq=pCq, then