Jtdylktuy

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

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

Hint:

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

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

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

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

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

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.

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.

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 $$frac{(x)(n)(1+x^4)^{n-1}(4x^3)-(1+x^4)^{n}(1)}{x^2}+frac{1}{x^2}=3binom{n}1 x^{2}+7binom{n}2 x^{6}+dots+(4n-1)binom{n}n x^{4n-2}$$ Substituting x=1 in the above equation $$3binom{n}1+7binom{n}2+dots+(4n-1)binom{n}n=(2n-1)2^n+1$$