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 $(^{20}_{14})5^5$
 A: 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 $$5^5 \binom{20}{6}.$$
A: From you try you get that the coefficient of $x^5$ is the sum of the coefficient of $x^5$ of each term of the sum, so it is :
$$\sum_{k=5}^{19} 5^5\binom{k}{5}=5^5\sum_{k=5}^{19} \binom{k}{5}$$
Then you can prove by induction that : $$\sum_{j=k}^n{j \choose k} = {n+1 \choose k+1} $$
It is the Hockey stick equality.
So the coefficient of $x^5$ in your expression is $5^5{20 \choose 6}$.
A: I know where you're stuck (I know this is from Allen's score series, since you are from India and preparing for IIT-JEE) :
Since $$\binom{n}{r}=\binom{n}{n-r}$$
Therefore :
$$\binom{20}{14}=\binom{20}{20-14}=\binom{20}{6}$$
And , since  :
$$\binom{n}{r}+\binom{n}{r-1}=\binom{n+1}{r}$$
$$\binom{20}{6}+\binom{20}{7}=\binom{21}{7} \implies \binom{20}{6}=\binom{21}{7}-\binom{20}{7} $$
