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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 ...

For r=0,1,2,........n,r = 0,1,2,........n, Prove that C0⋅Cr+C1⋅Cr+1+C2⋅Cr+2+....+Cn−r⋅Cn=2nC(n+r){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
A. C02+C12+C22+....+Cn2=2nCnC_0^2 + C_1^2 + C_{_2}^2 + .... + C_n^2{ = ^{2n}}{C_n}
B. C0⋅Cr+C1⋅Cr+1+C2⋅Cr+2+....+Cn−r⋅Cn=2nC(n+r){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)}}

Explanation

Solution

Hint : Here, we need to prove the given problem by using binomial co-efficient in the expansion of (1+x)n{(1 + x)^n} and (x+1)n{(x + 1)^n} . Each expansion has one more term than the power on the binomial. The sum of the exponents in each term in the expansion is the same as the power on the binomial.

Complete step by step solution:
We know that,
The binomial coefficient in the expansion,

C0+C1x+C2x2+......+Cnxn=(1+x)n→(1) C0xn+C1x+C2xn−2+....+Cn=(x+1)n→(2)   {C_0} + {C_1}x + {C_2}{x^2} + ...... + {C_n}{x^n} = {(1 + x)^n} \to (1) \\\ {C_0}{x^n} + {C_1}x + {C_2}{x^{n - 2}} + .... + {C_n} = {(x + 1)^n} \to (2) \;

By multiplying the above two series represent the coefficient of xn{x^n} (1)(1) and (2)(2)
(C0+C1x+C2x2+......+Cnxn)(C0xn+C1x+C2xn−2+....+Cn)=(1+x)2n→(3)({C_0} + {C_1}x + {C_2}{x^2} + ...... + {C_n}{x^n})({C_0}{x^n} + {C_1}x + {C_2}{x^{n - 2}} + .... + {C_n}) = {(1 + x)^{2n}} \to (3)
Equating coefficient of xn−r{x^{n - r}} from both sides of the equation (3)(3)
C0⋅Cr+C1⋅Cr+1+C2⋅Cr+2+....+Cn−r⋅Cn=2nC(n+r)→(4){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)}} \to (4)
Now put the value r=0r = 0 in equation (4)(4) , we get
C0⋅C0+C1⋅Cr+1+C2⋅C2+....+Cn⋅Cn=2nCn{C_0} \cdot {C_0} + {C_1} \cdot {C_{r + 1}} + {C_2} \cdot {C_2} + .... + {C_n} \cdot {C_n}{ = ^{2n}}{C_n}
Then, substituting r=1r = 1 in equation (4)(4) , we get
C0⋅C1+C1⋅C1+1+C2⋅C1+2+....+Cn−1⋅Cn=2nC(n+1){C_0} \cdot {C_1} + {C_1} \cdot {C_{1 + 1}} + {C_2} \cdot {C_{1 + 2}} + .... + {C_{n - 1}} \cdot {C_n}{ = ^{2n}}{C_{(n + 1)}}
By simplify the above binomial expression, we get
C0⋅C1+C1⋅C2+C2⋅C1+2+....+Cn−1⋅Cn=2nC(n+1){C_0} \cdot {C_1} + {C_1} \cdot {C_2} + {C_2} \cdot {C_{1 + 2}} + .... + {C_{n - 1}} \cdot {C_n}{ = ^{2n}}{C_{(n + 1)}}
Where, r=1r = 1
Therefore, C0⋅Cr+C1⋅Cr+1+C2⋅Cr+2+....+Cn−r⋅Cn=2nC(n+r){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)}}
Hence proved

Note : Each expansion has one more term than the power on the binomial. The sum of the exponents in each term in the expansion is the same as the power on the binomial. The powers on a in the expansion decrease by one with each successive term, while the powers on b increase by one. The coefficients form a symmetrical pattern.