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Question

For $$r = 0, 1, 2, ...., n$$, prove that $$C_{0}\cdot C_{r} + C_{1}\cdot C_{r + 1} + C_{2} \cdot C_{r + 2} + .... + C_{n - r} \cdot C_{n} = \ ^{2n}C_{(n + r)}$$ and hence deduce that
i) $$C_{0}^{2} + C_{1}^{2} + C_{2}^{2} + ...... + C_{n}^{2} = \ ^{2n}C_{n}$$
ii) $$C_{0}\cdot C_{1} + C_{1}\cdot C_{2} + C_{2}\cdot C_{3} + ..... + C_{n - 1} \cdot C_{n} = \ ^{2n}C_{n + 1}$$


Solution

To prove:
$$C_0.C_r+C_1.C_{r+1}+C_2.C_{r+2}+......+C_{n-r}.C_n={}^{2n}{C}_{n+r}$$
$$i)C_0^2+C_1^2+C_2^2+........+C_n^2={}^{2n}{C}_{n}$$
$$ii)C_0.C_1+C_1.C_{2}+C_2.C_{3}+......+C_{n-1}.C_n={}^{2n}{C}_{n+1}$$
Solution:
We know that,
$$C_0+C_1x+C_2x^2+......+C_nx^n=(1+x)^n$$         $$...............(1)$$
$$C_0x^n+C_1x^{n-1}+C_2x^{n-2}+......+C_n=(x+1)^n$$         $$...............(2)$$
Multiplying eqn.$$(1)$$ and eqn.$$(2)$$ we get,
$$(C_0+C_1x+C_2x^2+......+C_nx^n)$$ $$(C_0x^n+C_1x^{n-1}+C_2x^{n-2}+......+C_n)=(1+x)^{2n}$$            $$.............(3)$$
Equating coeffiecients of $$x^{n-r}$$ from both sides of $$(3)$$ we get,
$$C_0.C_r+C_1.C_{r+1}+C_2.C_{r+2}+......+C_{n-r}.C_n={}^{2n}{C}_{n+r}$$     $$..........(4)$$
Now putting $$r=0$$ in eqn.$$(4)$$ we get,
$$C_0.C_0+C_1.C_{1}+C_2.C_{2}+......+C_{n}.C_n={}^{2n}{C}_{n}$$
or, $$C_0^2+C_1^2+C_2^2+........+C_n^2={}^{2n}{C}_{n}$$
Now putting $$r=1$$ in eqn.$$(4)$$ we get,
$$C_0.C_1+C_1.C_{2}+C_2.C_{3}+......+C_{n-1}.C_n={}^{2n}{C}_{n+1}$$

Mathematics

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