c0,c1,c2 denotes coefficents expansion of (1 + x)^n , then c1 + c1c2 + c2c3 + .......cn - 1cn = (2n)!/(n + 1)!(n - 1)!
Answers 1
Step-by-step explanation:
We know, from definition of binomial expansion,
[tex]\sf \: {(1 + x)}^{n} = C_0 + C_1x + C_2 {x}^{2} + ... + C_{n - 1} {x}^{n - 1} + C_n {x}^{n} \\ \\ [/tex]
can also be expanded as
[tex]\sf \: {(1 + x)}^{n} = C_n + C_{n - 1}x + C_{n - 2} {x}^{2} + ... + C_{1} {x}^{n - 1} + C_0{x}^{n} \\ \\ [/tex]
On multiply above two equations, we get
[tex]\sf \: {(1 + x)}^{2n} =(C_0 + C_1x + C_2 {x}^{2} + ... + C_{n - 1} {x}^{n - 1} + C_n {x}^{n})( C_n + C_{n - 1}x + C_{n - 2} {x}^{2} + ... + C_{1} {x}^{n - 1} + C_0{x}^{n}) \\ \\ [/tex]
As we have to find the value of
[tex]\boxed{ \sf{ \:\sf \: \:C_1 + C_1C_2 + ... + C_{n - 1}C_n \: }} \\ \\ [/tex]
Now, On comparing the coefficient of [tex]x^{n-1} [/tex] on both sides, we get
[tex]\sf \: \: (^{2n}C_{n - 1}) =C_0C_1 + C_1C_2 + ... + C_{n - 1}C_n \\ \\ [/tex]
[tex]\sf \: \: \dfrac{{2n}!}{(n - 1)!(2n - n + 1)!} =C_0C_1 + C_1C_2 + ... + C_{n - 1}C_n \\ \\ [/tex]
[tex]\sf \: \: \dfrac{{2n}!}{(n - 1)!(n + 1)!} =C_0C_1 + C_1C_2 + ... + C_{n - 1}C_n \\ \\ [/tex]
can be rewritten as
[tex]\sf \: \:C_1 + C_1C_2 + ... + C_{n - 1}C_n = \dfrac{{2n}!}{(n - 1)!(n + 1)!} \\ \\ [/tex]
[tex]\qquad\boxed{ \sf{ \: \because \: C_0 = 1 \: }} \\ \\ [/tex]
Hence,
[tex]\sf\implies \sf \: \:C_1 + C_1C_2 + ... + C_{n - 1}C_n = \dfrac{{2n}!}{(n - 1)!(n + 1)!} \\ \\ [/tex]
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irene62
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