Question
Question: In the expansion of \({{\left( {{x}^{3}}-\dfrac{1}{{{x}^{2}}} \right)}^{n}},n\in N\), if the sum of ...
In the expansion of (x3−x21)n,n∈N, if the sum of the coefficient of x5 and x10 is 0, then the value of n is:
(a) 25
(b) 20
(c) 15
(d) None of these
Solution
Hint:Here, first we have to expand (x3−x21)n,n∈N with the help of binomial theorem, and then write the terms of x5 and x10, and find two equations for r and r’. Then, equate the sum of the coefficients of x5 and x10 to zero and obtain an equation for n in terms of r and r’. From the equations of r and r’ we will get the value of n.
Complete step-by-step answer:
Here, we are given (x3−x21)n,n∈N and also given that the sum of the coefficient of x5 and x10 is 0.
Now, we have to find the value of n.
We know by binomial theorem that:
(a+b)n=an+nC1an−1b+nC2an−2b2+.....+nCn−1abn−1+bn
The general term can be written as,
nCran−rbr
Now, consider (x3−x21)n,n∈N, by binomial expansion we get:
(x3−x21)n=(x3)n+nC1(x3)n(x21)+......+nCn−1x3(x21)n−1+(x21)n
Now, let us find the term of x5, we can take,
Term of x5=nCrx3r(−x21)n−r
We know that (ba)m=bmam
Term of x5=nCrx3r(x2)n−r(−1)n−r
We also know that (am)n=amn
Hence, we can write:
Term of x5=nCrx3rx2(n−r)(−1)n−r
⇒ Term of x5=nCrx3rx2n−2r(−1)n−r
We also have that am1=a−m.
Therefore, we can say that:
Term of x5=nCrx3rx−(2n−2r)(−1)n−r
⇒ Term of x5=nCrx3rx2r−2n(−1)n−r
By applying xmxn=xm+n,
⇒ Term of x5=nCrx3r+2r−2n(−1)n−r
Now, by equating the power of x on both the sides we get:
5=3r+2r−2n⇒5=5r−2n
Now, by taking -2n to the left side, we get:
5+2n=5r
Next by cross multiplication, we obtain:
r=55+2n …… (1)
Now, let us consider x10 be (r′+1)th term. Then we can say that,
Term of x10=nCr′x3r′(−x21)n−r′
Now, again by the same procedure, above, we will get:
⇒ Term of x10=nCr′x3r′+2r′−2n(−1)n−r′
Hence, by equating the power of x on both the sides we get:
10=3r′+2r′−2n⇒10=5r′−2n
Now, by taking -2n to the left side, we get:
10+2n=5r′
Next by cross multiplication, we obtain:
r′=510+2n …… (2)
We are given that the sum of the coefficients of x5 and x10 is zero.
Therefore, let us add the coefficients of x5 and x10, we obtain:
nCr(−1)n−r+nCr′(−1)n−r′=0
Next, by taking nCr′(−1)n−r′ to the right side we get:
nCr(−1)n−r=−nCr′(−1)n−r′
By taking modulus on both the sides we get:
nCr(−1)n−r=−nCr′(−1)n−r′⇒∣nCr∣=∣nCr′∣⇒nCr=nCr′
The above condition is possible only if r=r′or r=n−r′.
But r = r’ is not possible. Therefore, we can say that,