List I contains the different sum of the series and List II contains the maximum value of these sums. Match the lists.

a. ∑r=014 (15−r)r+115Cr20Cm−r             =∑r=014 (15−r)r+115!r!(15−r)!20Cm−r=∑r=014 15!(r+1)!(14−r)!20Cm−r=∑r=014 15Cr+120Cm−r=35Cm+1Which has maximum value  35C18.b.  35C0−35C1+35C2−35C3+…+(−1)r35Cr+…               = Coefficient of xr in 35C0−35C1x+35C2x2−35C3x3+…+(−1)r35Cr+…×1+x+x2+x3+…+xr+…             = Coefficient of xr in (1−x)35(1−x)−1= Coefficient of xr in (1−x)34=(−1)r34CrRequired maximum value is  34C17.c.  31Cr−5+5×31Cr−4+10×31Cr−3+10×31Cr−2+5×31Cr−1+31Cr             =5C5×31Cr−5+5C4×31Cr−4+5C3×31Cr−3+5C2×31Cr−2+5C1×31Cr−1+5C0×31Cr = Coefficient of xr in (1+x)5(1+x)31= Coefficient of xr in (1+x)36=36CrWhich has maximum value of  36C18.d. ∑r=0k 2k−r⋅(−1)r⋅37Cr⋅37−rC37−k           =∑r=0k 2k−r⋅(−1)r⋅37!(37−r)!r!(37−r)!(37−k)!(k−r)!=37!k!(37−k)!∑r=0k 2k−r⋅(−1)r⋅k!r!(k−r)!=37Ck∑r=0k kCr2k−r⋅(−1)r=37Ck(2−1)k=37CkWhich has maximum value  37C18.