Coefficient of $x^5$ in binomial expansion. We have to find coefficient of $x^5$ in the expansion of $$(1+5x)^5 +(1+5x)^6 +\cdots+(1+5x)^{19}$$
My try
from first expansion I get $(^5 _5)5^5$ from second $(^6 _5)5^5$ .
but from this I can prove the answer as $(^{20}_{14})5^5$

Better to first factor the expression and recognize it as the sum of a finite geometric series with common ratio $r = 1+5x$; thus $$\sum_{k=5}^{19} (1+5x)^k = (1+5x)^5 \sum_{k=0}^{14} (1+5x)^k = (1+5x)^5 \frac{(1+5x)^{15} - 1}{(1+5x) - 1} = \frac{(1+5x)^5 ((1+5x)^{15} - 1)}{5x}.$$  Letting $y = 5x$ and further expanding we get $$\frac{(1+y)^{20}}{y} - \frac{(1+y)^5}{y},$$ and if we want the coefficient of $x^5$, this corresponds to finding the coefficient of $y^5$ and multiplying by $5^5$.  The second term contributes no terms of this form, as the highest power is $y^4$.  Thus we are looking for the coefficient of $y^6$ in $(1+y)^{20}$, which is simply $\binom{20}{6} = \binom{20}{14}$, hence the coefficient of $x^5$ in the original expression is $$5^5 \binom{20}{6}.$$
