Find the best tutors and institutes for Class 11 Tuition
Search in
Find the 13th term in the expansion of
.
It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by 
.
Thus, 13th term in the expansion of 
is
Find the coefficient of x5 in
It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by 
.
Assuming that x5 occurs in the (r + 1)th term of the expansion (x + 3)8, we obtain
Comparing the indices of x in x5 and in Tr +1, we obtain
r = 3
Thus, the coefficient of x5 is
Find the coefficient of a5b7 in
It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by 
.
Assuming that a5b7 occurs in the (r + 1)th term of the expansion (a – 2b)12, we obtain
Comparing the indices of a and b in a5 b7 and in Tr +1, we obtain
r = 7
Thus, the coefficient of a5b7 is 
Write the general term in the expansion of
It is known that the general term Tr+1 {which is the (r + 1)th term} in the binomial expansion of (a + b)n is given by 
.
Thus, the general term in the expansion of (x2 – y6) is
Write the general term in the expansion of (x2 – yx)12, x ≠ 0
It is known that the general term Tr+1 {which is the (r + 1)th term} in the binomial expansion of (a + b)n is given by 
.
Thus, the general term in the expansion of(x2 – yx)12 is
Find the middle terms in the expansions of 
t is known that in the expansion of (a + b)n, if n is odd, then there are two middle terms, namely, 
term and 
term.
Therefore, the middle terms in the expansion of are 
term and 
term
Thus, the middle terms in the expansion of are 
.
Find the middle terms in the expansions of 
In the expansion of (1 + a)m + n, prove that coefficients of am and an are equal.
It is known that in the expansion (a + b)n, if n is even, then the middle term is 
term.
Therefore, the middle term in the expansion of is 
term
Thus, the middle term in the expansion of is 61236 x5y5.
The coefficients of the (r – 1)th, rth and (r + 1)th terms in the expansion of
(x + 1)n are in the ratio 1:3:5. Find n and r.
t is known that (k + 1)th term, (Tk+1), in the binomial expansion of (a + b)n is given by 
.
Therefore, (r – 1)th term in the expansion of (x + 1)n is 
r th term in the expansion of (x + 1)n is 
(r + 1)th term in the expansion of (x + 1)n is 
Therefore, the coefficients of the (r – 1)th, rth, and (r + 1)th terms in the expansion of (x + 1)n are 
respectively. Since these coefficients are in the ratio 1:3:5, we obtain
Multiplying (1) by 3 and subtracting it from (2), we obtain
4r – 12 = 0
⇒ r = 3
Putting the value of r in (1), we obtain
n – 12 + 5 = 0
⇒ n = 7
Thus, n = 7 and r = 3
Prove that the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1 .
It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by 
.
Assuming that xn occurs in the (r + 1)th term of the expansion of (1 + x)2n, we obtain
Comparing the indices of x in xn and in Tr + 1, we obtain
r = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n is
Assuming that xn occurs in the (k +1)th term of the expansion (1 + x)2n – 1, we obtain
Comparing the indices of x in xn and Tk + 1, we obtain
k = n
Therefore, the coefficient of xn in the expansion of (1 + x)2n –1 is
From (1) and (2), it is observed that
Therefore, the coefficient of xn in the expansion of (1 + x)2n is twice the coefficient of xn in the expansion of (1 + x)2n–1.
Hence, proved.
Write the general term in the expansion of (x2 – yx)12, x ? 0
It is known that the general term Tr+1 {which is the (r + 1)th term} in the binomial expansion of (a + b)n is given by .
Thus, the general term in the expansion of(x2 – yx)12 is
Find a positive value of m for which the coefficient of x2 in the expansion(1 + x)m is 6.
It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by 
.
Assuming that x2 occurs in the (r + 1)th term of the expansion (1 +x)m, we obtain
Comparing the indices of x in x2 and in Tr + 1, we obtain
r = 2
Therefore, the coefficient of x2 is
.
It is given that the coefficient of x2 in the expansion (1 + x)m is 6.
Thus, the positive value of m, for which the coefficient of x2 in the expansion
(1 + x)m is 6, is 4.
Find the 4th term in the expansion of (x – 2y)12 .
It is known that (r + 1)th term, (Tr+1), in the binomial expansion of (a + b)n is given by 
.
Thus, the 4th term in the expansion of (x – 2y)12 is
How helpful was it?
How can we Improve it?
Please tell us how it changed your life *
Please enter your feedback
UrbanPro.com helps you to connect with the best Class 11 Tuition in India. Post Your Requirement today and get connected.
Find best tutors for Class 11 Tuition Classes by posting a requirement.
Get started now, by booking a Free Demo Class
