Coefficient of $x^5$ in $(1+x)^{21} + (1+x)^{22} + ... + (1+x)^{30}$

Question

How do I find the coefficient of $x^5$ in the expansion of $(1+x)^{21} + (1+x)^{22} + ... + (1+x)^{30}$?

Answer 1

Use binomial theorem we have : the coefficient of $x^5$ in $(1+x)^n$ is $\binom{n}{5}$. Then we have:

$$\sum_{i=21}^{30}\binom{i}{5}=682017$$

Answer 2

We have

$(1+x)^n=\sum_{k=0}^{n}\binom{n}{k}x^k$,

hence the coefficient of $x^5$ is $\binom{n}{5}$.

Threrfore the coefficient you are looking for is given by

$\binom{21}{5}+\binom{22}{5}+...+\binom{30}{5}$

Answer 3

$$\sum_{n=21}^{30}(1+x)^n=(1+x)^{21}\cdot\dfrac{(1+x)^{9+1}-1}{(1+x)-1}=\dfrac{(1+x)^{21}\{(1+x)^{10}-1\}}x$$

So, we need the coefficient of $x^6$ in $$(1+x)^{21}\{(1+x)^{10}-1\}=(1+x)^{31}-(1+x)^{10}$$

which will be $$\binom{31}6-\binom{21}6$$

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Alternatively, the coefficient of $x^5$ will be

$$\sum_{n=21}^{30}\binom n5$$

We can use the Pascal's rule to equate the two seemingly different results.