Marching band conductor

Let $f(x) = a_0+a_1 x + ... + a_k x^k$ . Then $f(n) = a_0 + ... + a_kn^k$ and $f(n-1)=a_0+...+a_k(n-1)^k$ . $f(n)-f(n-1)=n^{101}=a_kkn^{k-1}+...$ . So we have $k-1=101$ and $a_kk=1 \implies a_k={1 \over 102}$ .

$f(n) = n^{101} + f(n-1), f(1)=1$ . We shall assume f(x) is a polynomial of degree 102. It follows that $\sum_{i=0}^{102} a_{i}n^{i} = n^{101}+\sum_{i=0}^{102} a_{i}(n-1)^{i}$ $\sum_{i=0}^{102} a_{i}(n^{i}-(n-1)^{i}) = n^{101}$ By the difference of powers formula $n^{101}$ $=\sum_{i=0}^{102} a_{i}(n^{i}-(n^{i}-1))(n^{i-1}+n^{i-2}(n-1)+...+n(n-1)^{i-2}+(n-1)^{i-1})$ $=\sum_{i=0}^{102} a_{i}(n^{i-1}+n^{i-2}(n-1)+...+n(n-1)^{i-2}+(n-1)^{i-1})$ In the i=102 term of the summation, we have a polynomial of degree 101 with a leading term of $a_{i}102n^{101}$ . As this is the only source of $n^{101}$ in the summation, $a_{i}=\frac{1}{102}$

Note that $\sum_{i = 1}^n \bigg((i + 1)^{102} - i^{102}\bigg) = (n + 1)^{102} - 1$ i.e. $\sum_{i = 1}^n \bigg(\binom{102}{1} i^{101} + \dots \bigg) = n^{102} + \dots$ where the dots don't denote infinite sum but sum of terms which we don't care about, now just divide the equation by $\binom{102}{1} = 102$ we get leading coefficient $$\frac{1}{102}$$ Hence $$a + b = \boxed{103}$$

use property of converting summation into integral u will get (n^102)/102 so ans is 1+102