Sum of Some Digits

We know by Binomial Theorem

$(x+1)^m = x^m + mC_1\cdot x^{m-1} + mC_2\cdot x^{m-2} + \cdots + mC_{m-2}\cdot x^2 + mC_{m-1}\cdot x + 1 \\ hence, \\ ((a\times10^n)+1)^m=(a\times10^n)^m + mC_1\cdot (a\times10^n)^{m-1} + mC_2\cdot (a\times10^n)^{m-2} + \cdots + mC_{m-2}\cdot (a\times10^n)^2 + mC_{m-1}\cdot (a\times10^n) + 1 \\ = A +mC_{m-2}\cdot (a\times10^n)^2+ mC_{m-1}\cdot (a\times10^n) + 1 \text{ where, A represents remaining terms}\\ =A +\frac{m(m-1)}{2}\cdot (a\times10^n)^2+ m\cdot a\times10^n + 1$

In our case:

$a=99, n=11, m=987654321 \\ (99\times10^{11}+1)^{987654321}=A+\frac{987654321\cdot987654322}{2}\times10^{22}+987654321\cdot99\times10^{11}+1 \\ or, \\ (99\times10^{11}+1)^{987654321}=A+\frac{987654321\cdot987654322}{2}\times10^{22}+97777777779\times10^{11}+1$

From this it is clear that last 22 digits are:

$97777777779\times10^{11}+1$

hence, the needed sum = $82$

Mathematica

Tr@IntegerDigits@PowerMod[9*10^11+1,987654321,10^22]

returns 82