How can we evaluate the sequence ${}_3C_1+{}_7C_2+{}_{11}C_3+\dots+{}_{4n-1}C_n=$?

Question

I am new in combinatorics. How is the following sequence calculated?

${}_3C_1+{}_7C_2+{}_{11}C_3+\dots+{}_{4n-1}C_n=$?

Answer 1

Hint:

The $r$th term$T_{r+1}$ of $(x^a+x^b)^{4m-1}$ is $\displaystyle\binom{4m-1}r x^{a(4m-1-r)+br}$

$r=m\implies T_{m+1}=\displaystyle\binom{4m-1}m x^{a(4m-1-m)+bm}=\binom{4m-1}m x^{m(3a+b)-a}$

If we set $3a+b=0\iff b=-3a$

$T_{m+1}$ of $(x^a+x^{-3a})^{4m-1}$ will be $\displaystyle\binom{4m-1}m x^{-a}$

Set $a=-1,T_{r+1}$ in $(x^{-1}+x^3)^{4m-1}$ will be $\displaystyle\binom{4m-1}mx$

So, we need to find coefficient of $x$ in $$\sum_{m=1}^n(x^{-1}+x^3)^{4m-1}$$ which is a finite Geometric Series.

Answer 2

Here are some java code to compute it.

public static BigInteger ComputeSerie(int x){
            BigInteger sum=BigInteger.ZERO;
            for (int i=1;i<=x;i++)
                sum=sum.add(binomial(4*i-1,i));
            return sum;
        }
        public static BigInteger binomial(final int N, final int K) {
            BigInteger ret = BigInteger.ONE;
            for (int k = 0; k < K; k++) {
                ret = ret.multiply(BigInteger.valueOf(N-k))
                         .divide(BigInteger.valueOf(k+1));
            }
            return ret;
        }
System.out.println(ComputeSerie(20)) gives 2973189430714766571.

Answer 3

Consider expansion of $$(1+x^4)^{n}=\binom{n}0 x^{0}+\binom{n}1 x^{4}+\binom{n}2 x^{8}+\dots+\binom{n}n x^{4n}$$ $$(1+x^4)^{n}-\binom{n}0x^{0}=\binom{n}1 x^{4}+\binom{n}2 x^{8}+\dots+\binom{n}n x^{4n} $$Dividing by $x$ on both sides $$\frac{(1+x^4)^{n}-1}{x}=\binom{n}1 x^{3}+\binom{n}2 x^{7}+\dots+\binom{n}n x^{4n-1}$$ Differentiating w.r.t x on both sides $$\frac{(x)(n)(1+x^4)^{n-1}(4x^3)-(1+x^4)^{n}(1)}{x^2}+\frac{1}{x^2}=3\binom{n}1 x^{2}+7\binom{n}2 x^{6}+\dots+(4n-1)\binom{n}n x^{4n-2}$$ Substituting x=1 in the above equation $$3\binom{n}1+7\binom{n}2+\dots+(4n-1)\binom{n}n=(2n-1)2^n+1$$