If the sum of the coefficients in the expansion of $(p^{2} x^{2} - 2 p x + 1)^{51}$ vanishes, then $p =$

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Solution

The correct option is B $1$
Putting $x = 1$ , we get
$(p^{2} - 2 p + 1)^{51}$
$= ((p - 1)^{2})^{51}$
$= (p - 1)^{102}$
$= (1 - p)^{102}$
Expanding using binomial theorem, we get
$^{102} C_{0} -^{102} C_{1} (p) +^{102} C_{2} (p^{2}) -^{102} C_{3} (p^{3}) . . .^{102} C_{102} (p^{102})$
Therefore the sum of coefficients are
$^{102} C_{0} -^{102} C_{1} +^{102} C_{2} -^{102} C_{3} . . .^{102} C_{102}$
$= 1 -^{102} C_{1} +^{102} C_{2} -^{102} C_{3} . . .^{102} C_{102}$
$= 1 + [^{102} C_{2} +^{102} C_{4} +^{102} C_{6} . . .^{102} C_{102}] - [^{102} C_{1} +^{102} C_{3} +^{102} C_{3} . . .^{102} C_{101}]$
$= 1 + 2^{n - 1} - 2^{n - 1}$
$= 1$
Hence answer is C

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