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Question

Mathematics Question on Binomial theorem

The coefficient of x70x^{70} in x2(1+x)98+x3(1+x)97+x4(1+x)96++x54(1+x)46x^2 (1+x)^{98} + x^3 (1+x)^{97} + x^4 (1+x)^{96} + \ldots + x^{54} (1+x)^{46} is 99Cp46Cq^{99}C_p - ^{46}C_q.
Then a possible value to p+qp + q is:

A

55

B

61

C

68

D

83

Answer

83

Explanation

Solution

x2(1+x)98+x3(1+x)97+x4(1+x)96++x54(1+x)46x^2(1+x)^{98} + x^3(1+x)^{97} + x^4(1+x)^{96} + …… + x^{54}(1+x)^{46}

The coefficient of x70x^{70} is:

98C68+97C67+96C66+^{98}C_{68} + ^{97}C_{67} + ^{96}C_{66} + \cdots

Simplify:

47C17+46C16^{47}C_{17} + ^{46}C_{16}

Combine terms:

46C30+46C31+47C30+{^{46}}C_{30} + {^{46}}C_{31} + ^{47}C_30 + \cdots

Using binomial expansion:

47C30+=99Cp46Cq{^{47}}C_{30} + \cdots = ^{99}C_p - ^{46}C_q

Possible values of p+qp+q:

p+q=62,83,99,46p+q = 62, 83, 99, 46

Final Answer:

p+q=83Option (4).p+q = 83 \quad \text{Option (4)}.