What makes a number binomial?: Identity 1

Probability Level 2

Find the last two digits of ∑ n = 0 15 ( 15 n ) \large\sum_{n = 0}^{15}\binom{15}{n}


The answer is 68.

Sanchayapol Lewgasamsarn by Sanchayapol Lewgasamsarn , 20, Thailand

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1 solution

Note that the question can be rewritten as

( 15 0 ) + ( 15 1 ) + ( 15 2 ) + . . . + ( 15 15 ) \binom{15}{0} + \binom{15}{1} + \binom{15}{2} + ... + \binom{15}{15}

By observing and the binomial theorem,

( 15 0 ) + ( 15 1 ) + ( 15 2 ) + . . . + ( 15 15 ) = ( 1 + 1 ) 15 \binom{15}{0} + \binom{15}{1} + \binom{15}{2} + ... + \binom{15}{15} = (1 + 1)^{15}

Thus, ( 15 0 ) + ( 15 1 ) + ( 15 2 ) + . . . + ( 15 15 ) = 2 15 = 32768 \binom{15}{0} + \binom{15}{1} + \binom{15}{2} + ... + \binom{15}{15} = 2^{15} = 32768

Thus, the last two digits are 68 \boxed{68}

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