If C0-C1-C2- are combinational coefficient in the expansion of 1-xn- n- and C0-C1-C1-C2-C2-C3-C19-C20-C0-C1-C2-C18-a20b- a-b- then the value of a-b is

L.H.S =C_0+C_1)·(C_1+C_2)· (C_2+C_3)⋯(C_19+C_20) Here C_r= ^20C_r We know, ^nC_r+ ^nC_r 1= ^n+1C_r ⇒ C_0+C_1= ^20+1C_1= ^21C_1, C_1+C_2= ^21C_2,⋯, C_19+C_20= ^2 ...

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Byju's Answer

Standard XII

Mathematics

Division and Distributuion into Groups of Unequal Sizes.

If C0,C1,C2⋯ ...

Question

If $C_{0}, C_{1}, C_{2} \dots$ are combinational coefficient in the expansion of $(1 + x)^{n}; n \in N$ and $(C_{0} + C_{1}) \cdot (C_{1} + C_{2}) \cdot (C_{2} + C_{3}) \dots (C_{19} + C_{20}) = \frac{C_{0} \cdot C_{1} \cdot C_{2} \dots C_{18} \cdot a^{20}}{b!}; (a, b \in N)$ , then the value of $(a + b)$ is

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Solution

$L.H.S = C_{0} + C_{1}) \cdot (C_{1} + C_{2}) \cdot (C_{2} + C_{3}) \dots (C_{19} + C_{20})$
Here $C_{r} =^{20} C_{r}$
We know,
$^{n} C_{r} +^{n} C_{r - 1} =^{n + 1} C_{r}$
$\Rightarrow C_{0} + C_{1} =^{20 + 1} C_{1} =^{21} C_{1}, C_{1} + C_{2} =^{21} C_{2}, \dots, C_{19} + C_{20} =^{21} C_{20}$
Putting these values in $L.H.S$
$L.H.S =^{21} C_{1} \cdot^{21} C_{2} \cdot^{21} C_{3} \dots^{21} C_{20}$
also $^{n + 1} C_{r} = \frac{n + 1}{r}^{n} C_{r - 1}$

$L.H.S = \frac{21}{1} \cdot^{20} C_{0} \cdot \frac{21}{2} \cdot^{20} C_{1} \cdot \dots \frac{21}{20}^{20} C_{19}$
$= \frac{C_{0} \cdot C_{1} \cdot C_{2} \dots C_{19} \cdot 21^{20}}{20!}$
$= \frac{C_{0} \cdot C_{1} \cdot C_{2} \dots C_{18} \cdot 20 \cdot 21^{20}}{20 \cdot 19!}$
$= \frac{C_{0} \cdot C_{1} \cdot C_{2} \dots C_{18} \cdot 21^{20}}{19!}$
$\Rightarrow a = 21, b = 19$
$∴ a + b = 40$

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If C0,C1,C2⋯ are combinational coefficient in the expansion of (1+x)n; n∈N and (C0+C1)⋅(C1+C2)⋅(C2+C3)⋯(C19+C20)=C0⋅C1⋅C2⋯C18⋅a20b!; (a,b∈N), then the value of (a+b) is