find the coefficient of $x^{25}$

Question

find the coefficient of $x^{25}$ in $(x+ 2x^{2} + 3x^{3} +..) (x^{3}+x^{4}+x^{5}+..)^{6}$

when I'm taking $x$ out I'm getting a coefficient of $x^{19}$ .. Can someone give a hand please? thanks

Answer 1

After OP's edit.

Hint 1: It is equivalent to: $$[x^{25}](x+2x^2+3x^3+\cdots)(x^3+x^4+x^5+\cdots)^{\color{red}6}=\\ [x^{25}]x^{19}(1+2x+3x^2+\cdots)(1+x+x^2+\cdots)^6=\\ [x^6](1+2x+3x^2+\cdots)(1+x+x^2+\cdots)^6$$

Hint2: The pairs of powers of $x$ are: $$(0,6),(1,5),(2,4),...,(6,0)$$ Hint 3: $$[x^6](1+2x+3x^2+\cdots)\left(\frac1{1-x}\right)^6=\\ [x^6](1+2x+3x^2+\cdots)(1-x)^{-6}\stackrel{*}=\\ [x^6](1+2x+3x^2+\cdots)\sum_{i=0}^{\infty} {5+i\choose i}x^i=\\ \small{1\cdot {11\choose 6}+2\cdot {10\choose 5}+3\cdot {9\choose 4}+4\cdot {8\choose 3}+5\cdot {7\choose 2}+6\cdot {6\choose 1}+7\cdot {5\choose 0}}= 1716.$$ Note: At $\stackrel{*}=$, it was used Negative binomial theorem.

Answer 2

To systematically solve this question, you need to use the binomial series, which is an extension of the binomial theorem to negative integer powers. Starting with the geometric series, which you should recognize in the second factor, $$ (1-x)^{-1}=1+x+x^2+x^3+... $$ you get $$ (1-x)^{-d}=(1+x+x^2+x^3+...)^d=\sum_{n=0}^\infty\binom{n+d-1}{d-1}x^n \\ =1+dx+\frac{d(d+1)}2x^2+\frac{d(d+1)(d+2)}6x^3+... $$ Especially you can recognize that the first factor can be identified as $$ x+2x^2+3x^3+... = x(1-x)^{-2}. $$

Then the product is equal to $$ x^{1+6\cdot 3}(1-x)^{-2-6}=x^{19}\sum_{n=0}^\infty\binom{n+7}{7}x^n, $$ so that the coefficient you need to evaluate is $$ \binom{13}7=\binom{13}6=\frac{13⋅12⋅11⋅10⋅9⋅8}{1⋅2⋅3⋅4⋅5⋅6}=13⋅12⋅11=1716. $$

Answer 3

The sequence is $$S(x)=[(x+2x^2+5x^3+...)(x^3+x^5+x^4+...)^6]= \left(\frac{x}{(1-x)^2} \frac{x^{18}}{(1-x)^{6}}\right)=\left(\frac{x^{19}}{(1-x)^8}\right).$$ Then coefficient of $x^{25}$ in $S(x)$ is the co-efficient of $x^6$ in $(1-x)^{-8}$ which by infinite binomial series is ${-8 \choose 6}={13 \choose 6}.$ Note that $$(1+z)^{-n}=\sum_{k=0}^{\infty} {-n \choose k} ~z^k, ~|z|<1$$

$${-n \choose k}=(-1)^k {n+k-1 \choose k}$$

Answer 4

$$(x+ 2x^{2} + 3x^{3} +..) (x^{3}+x^{4}+x^{5}+..)^{6}$$ $$=x^{18}\dfrac{x}{(1-x)^2}\dfrac{1}{(1-x)^6}$$ $$=x^{19}\dfrac{1}{(1-x)^8}$$

The generating function $$\frac1{(1-x)^k}=\sum_\limits{n=k-1}^\infty \binom{n}{k-1}x^{n-k+1}$$

means we want $n-8+1=6$, and so $n=13$.

Therefore the answer is $\binom{13}{8-1}=1716$.

Answer 5

Using generating functions, the coefficient we’re interested in is $$[x^{25}]\left(x\frac d{dx}\frac1{1-x}\right)\left({x^3\over1-x}\right)^6 = [x^{25}]{x\over(1-x)^2}{x^{18}\over(1-x)^6} = [x^6]{1\over(1-x)^8} = \binom{13}6.$$ The last equality comes from the generalized Binomial Theorem.