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Question: Find the coefficient of \({a^2}{b^5}d\) in the expansion of \({\left( {a + b - c - d} \right)^8}\)....

Find the coefficient of a2b5d{a^2}{b^5}d in the expansion of (a+bcd)8{\left( {a + b - c - d} \right)^8}.

Explanation

Solution

Hint : In this question we will use the binomial theorem for the expansion of the given term. To find the coefficient of (r+1)th{\left( {r + 1} \right)^{th}}term of an expansion (a+b)n{\left( {a + b} \right)^n}, first we have to find that (r+1)th{\left( {r + 1} \right)^{th}} term ,which can be identified by the formulae :

Complete step-by-step answer:
Tr+1= nCranr.br{T_{r + 1}} = {{\text{ }}^n}{C_r}{a^{n - r}}.{b^r} . when we get that term , we will easily get the coefficient of that term.
Given that , we have to find the coefficient of a2b5d{a^2}{b^5}d
Given expansion is , (a+bcd)8{\left( {a + b - c - d} \right)^8}
According to the binomial theorem, we know that for an expansion ,
(a+b)n{\left( {a + b} \right)^n}, the general term can be given as : Tr+1= nCranr.br{T_{r + 1}} = {{\text{ }}^n}{C_r}{a^{n - r}}.{b^r}
In this question we have expansion (a+bcd)8{\left( {a + b - c - d} \right)^8}
So here, n=8n = 8
Let us suppose that the coefficient a2b5d{a^2}{b^5}d occurs in the (r+1)th{\left( {r + 1} \right)^{th}} term of the expansion (a+bcd)8{\left( {a + b - c - d} \right)^8}.
Now, we know that
Tr+1= nCranr.br{T_{r + 1}} = {{\text{ }}^n}{C_r}{a^{n - r}}.{b^r} …..(i)
By comparing the indices of equation (i) with (a+bcd)8{\left( {a + b - c - d} \right)^8} and using the equation,
We get the coefficient of a2b5d{a^2}{b^5}d as (a)2×(b)5×(c)0×(d)1=8!2! × 5! × 0! × 1!  (a)2×(b)5×(c)0×(d)1 = 8×7×6×5×4×3×2×12×1×5×4×3×2×1×1  (a)2×(b)5×(c)0×(d)1 = 8×7×3 (a)2×(b)5×(c)0×(d)1 = 168.  \Rightarrow {(a)^2} \times {(b)^5} \times {( - c)^0} \times {( - d)^1} = \dfrac{{8!}}{{2!{\text{ }} \times {\text{ }}5!{\text{ }} \times {\text{ }}0!{\text{ }} \times {\text{ }}1!}} \\\ \\\ \Rightarrow {(a)^2} \times {(b)^5} \times {( - c)^0} \times {( - d)^1}{\text{ = }}\dfrac{{8 \times 7 \times 6 \times 5 \times 4 \times 3 \times 2 \times 1}}{{2 \times 1 \times 5 \times 4 \times 3 \times 2 \times 1 \times 1}} \\\ \\\ \Rightarrow {(a)^2} \times {(b)^5} \times {( - c)^0} \times {( - d)^1}{\text{ = 8}} \times 7 \times 3 \\\ \Rightarrow {(a)^2} \times {(b)^5} \times {( - c)^0} \times {( - d)^1}{\text{ = }}168. \\\ .
Hence,the coefficient of a2b5d{a^2}{b^5}d in the expansion of (a+bcd)8{\left( {a + b - c - d} \right)^8} is : 168.
Note: In this type of question we will use the concept of binomial theorem for expansion. First we will get the general terms like r, n,a,b etc and then by putting these values in the formulae for getting the general term of binomial expansion i.e. Tr+1= nCranr.br{T_{r + 1}} = {{\text{ }}^n}{C_r}{a^{n - r}}.{b^r}. Here we can see that the term is Tr+1{T_{r + 1}} and its coefficient is  nCr{{\text{ }}^n}{C_r}. Through this we will get our answer .