Answer: Given, 23n – 7n – 1 23n – 7n – 1 = 8n – 7n – 1 Using binomial theorem, 8n = 7n + 1 8n = (1 + 7) n 8n = nC0 + nC1 (7)1 + nC2 (7)2 + nC3 (7)3 + nC4 (7)2 + nC5 (7)1 + … + nCn (7) n 8n = 1 + 7n...
The two successive terms in the expansion of whose coefficients are in the ratio 1: 4 are
(A) and
(B) and
(C) and
(D) and
(C) $5^{\text {th }}$ and $6^{\text {th }}$ Explanation: Let $(r+1)^{\text {th }}$ and $(r+$ 2 ) th term be the two successive terms in the expansion of $(1+x)^{24}$. Now, $T_{r+1}={ }^{24} C_{r}...
Given the integers , and coefficients of and terms in the binomial expansion of are equal, then
(B)
(C)
(D) none of these
(A) $n=2 r$ Explanation: Given expression is $(1+x)^{2 n}$ $T_{3 r}=T_{(3 r-1)+1}={ }^{2 n} C_{3 r-1} x^{3 r-1}$ $T_{r+2}=T_{(r+1)+1}={ }^{2 n} \mathrm{C}_{r+1} x^{r+1}$ ${ }^{2 n} C_{3 r-1}={ }^{2...
The total number of terms in the expansion of after simplification is
(A) 50
(B) 202
(C) 51
(D) none of these
(C) 51 Explanation: Given expression is $(x+a)^{100}+(x-a)^{100}$ $\begin{aligned} =&\left({ }^{100} C_{0} x^{100}+{ }^{100} C_{1} x^{99} a+{ }^{100} C_{2} x^{98} a^{2}+\ldots\right) \\...
Find the term independent of in the expansion of
Given expression is, $\left(1+x+2 x^{3}\right)\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ Considering $\left(\frac{3}{2} x^{2}-\frac{1}{3 x}\right)^{9}$ Using standard formula, we get...
If occurs in the expansion of . Prove that its coefficient is
Given expression is $\left(x^{2}+\frac{1}{x}\right)^{2 n}$ Using the standard formula, we get, $T_{r+1}={ }^{2 n} C_{r}\left(x^{2}\right)^{2 n-r}\left(\frac{1}{x}\right)^{r}={ }^{2 n} C_{r} x^{4 n-3...
In the expansion of if the sum of odd terms is denoted by and the sum of even term by . Then prove that
(i)
(ii)
(i) We know, $(x+a)^{n}=$ ${ }^{n}C_{0} x^{n}+{ }^{n} C_{1} x^{n-1} a^{1}+{ }^{n} C_{2} x^{n-2} a^{2}+{ }^{n} C_{3} x^{n-3} a^{3}+\ldots$ Sum of odd terms will be, $O={ }^{n} C_{0} x^{n}+{ }^{n}...
Find in the binomial , if the ratio of term from the beginning to the term from the end is .
Given expression is $\left(\sqrt[3]{2}+\frac{1}{\sqrt[3]{3}}\right)^{n}$ Now, the $7^{\text {th }}$ term from beginning will be, $T_{7}=T_{6+1}={ }^{n}...
Show that the middle term in the expansion of is
Given, expression is $\left(x-\frac{1}{x}\right)^{2 n}$. Since the index is given as $2 n$, which is even. So, there is only one middle term, i.e., $\left(\frac{2 n}{2}+1\right)$ th term $=(n+1)$ th...
If is a real number and if the middle term in the expansion of Is 1120 , find .
Given expansion is $\left(\frac{p}{2}+2\right)^{8}$ Since the index is given as $n=8$, there is only one middle term, i.e., $\left(\frac{8}{2}+1\right)$ th $=5^{\text {th }}$ term $T_{5}=T_{4+1}={...
Find the coefficient of in the expansion of
Given expression is $\left(1+x+x^{2}+x^{3}\right)^{11}$ $=\left[(1+x)+x^{2}(1+x)\right]^{11}=\left[(1+x)\left(1+x^{2}\right)\right]^{11}=(1+x)^{11} \cdot\left(1+x^{2}\right)^{11}$ $=\left({ }^{11}...
If the coefficient of second, third and fourth terms in the expansion of are in A.P. Show that .
Given function is $(1+x)^{2 n}$ ${ }^{2 n} C_{1},{ }^{2 n} C_{2}$ and ${ }^{2 n} C_{3}$ are the Coefficient of $2^{\text {nd }}, 3^{\text {rd }}$ and $4^{\text {th }}$ terms are respectively. ${...
Find the value of , if the coefficients of and terms in the expansion of are equal.
Given function is $(1+x)^{18}$ Now, $(2 r+4)^{\text {th }}$ term will be equal to $T_{(2 r+3)+1}$ $T_{(2 r+3)+1}={ }^{18} C_{2 r+3}(x)^{2 r+3}$ And $(r-2)^{\text {th }}$ term, that is $T_{(r-3)...
Find the sixth term of the expansion if the binomial coefficient of the third term from the end is
Given function is $\left(y^{1 / 2}+x^{1 / 3}\right)^{n}$ Given the binomial coefficient of the third phrase from the beginning, $=45$ So, ${ }^{n} C_{n-2}=45$ The above expression can be re-written...
Find the coefficient of in the expansion of
Given function is $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}$ We may write the provided expression as $T_{r+1}$ using the conventional formula, $T_{r+1}={ }^{15}...
Find the coefficient of in the expansion of
Given function is $\left(x-x^{2}\right)^{10}$ $T_{r+1}={ }^{10} C_{r} x^{10-r}\left(-x^{2}\right)^{r}=(-1)^{r 10} C_{r} x^{10-r} x^{2 r}=(-1)^{r 10} C_{r} x^{10+r}$ For the coefficient of $x^{15}$,...
Find the middle term (terms) in the expansion of
(i)
(ii)
(i) Given function is $\left(\frac{x}{a}-\frac{a}{x}\right)^{10}$ Here, Index of $n$ is $10$ which is even number. So, there is one middle term which is $(10 / 2+1)^{\text {th }}$ term that is...
Find the coefficient of in the expansion of
Given function is $\left(1-3 x+7 x^{2}\right)(1-x)^{16}$ Expansion of the function is, $=\left(1-3 x+7 x^{2}\right)\left({ }^{16} C_{0}-{ }^{16} C_{1} x^{1}+{ }^{16} C_{2} x^{2}+\ldots+{ }^{16}...
If the term free from in the expansion of is 405, find the value of k.
Given is, $\sqrt{x}-\frac{k}{x^{2}}$ From the standard formula of $T_{r+1}$ we know, $T_{r+1}={ }^{10} C_{r}(\sqrt{x})^{10-r}\left(\frac{-k}{x^{2}}\right)^{r}={ }^{10}...
Find the term independent of , in the expansion of
Given function is $\left(\frac{3 x^{2}}{2}-\frac{1}{3 x}\right)^{15}$ We know, the standard formula of $T_{r+1}$ will be, $T_{r+1}={ }^{15} C_{r}\left(\frac{3...
Find the expansion of (3×2 – 2ax + 3a2)3 using binomial theorem
We know that \[\left( a\text{ }+\text{ }b \right)3\text{ }=\text{ }a3\text{ }+\text{ }3a2b\text{ }+\text{ }3ab2\text{ }+\text{ }b3\] Putting\[a\text{ }=\text{ }3x2\text{ }\And \text{ }b\text{...
Expand using Binomial Theorem
Utilizing binomial hypothesis the given articulation can be extended as Again by utilizing binomial hypothesis to grow the above terms we get From condition 1, 2 and 3 we get
Find n, if the ratio of the fifth term from the beginning to the fifth term from the end in the expansion of NCERT Solutions for Class 11 Maths Chapter 8 Binomial Theorem Image 39 is √6: 1
Find an approximation of (0.99)5 using the first three terms of its expansion
\[0.99\] can be composed as \[0.99\text{ }=\text{ }1\text{ }\text{ }0.01\] Presently by applying binomial hypothesis we get \[\left( o.\text{ }99 \right)5\text{ }=\text{ }\left( 1\text{ }\text{...
Find the value of
Evaluate
Utilizing binomial hypothesis the articulation \[\left( a\text{ }+\text{ }b \right)6\] and\[\left( a\text{ }+\text{ }b \right)6\] , can be extended \[\left( a\text{ }+\text{ }b \right)6\text{...
If a and b are distinct integers, prove that a – b is a factor of an – bn, whenever n is a positive integer. [Hint write an = (a – b + b)n and expand]
To demonstrate that \[\left( a\text{ }\text{ }b \right)\] is a factor of\[\left( an\text{ }\text{ }bn \right)\] , it must be demonstrated that \[an\text{ }\text{ }bn\text{ }=\text{ }k\text{...
Find the coefficient of
x5 in the product (1 + 2x)6 (1 – x)7 using binomial theorem. Solution: (1 + 2x)6 = 6C0 + 6C1 (2x) + 6C2 (2x)2 + 6C3 (2x)3 + 6C4 (2x)4 + 6C5 (2x)5 + 6C6 (2x)6 = 1 + 6 (2x) + 15 (2x)2 + 20 (2x)3 + 15...
Find a if the coefficients of x2 and x3 in the expansion of (3 + a x)9 are equal
Find a, b and n in the expansion of (a + b)n if the first three terms of the expansion are 729, 7290 and 30375, respectively
We know that \[\left( r\text{ }+\text{ }1 \right)th\] term, \[\left( Tr+1 \right)\] , in the binomial expansion of \[\left( a\text{ }+\text{ }b \right)n\] is given by \[Tr+1\text{ }=\text{...
Find a positive value of m for which the coefficient of x2 in the expansion (1 + x)m is 6
The overall term \[Tr+1\] in the binomial extension is given by \[Tr+1\text{ }=\text{ }nCr\text{ }an-r\text{ }br\] Here\[a\text{ }=\text{ }1\] , \[b\text{ }=\text{ }x\] and \[n\text{ }=\text{ }m\]...
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
The overall term \[Tr+1\] in the binomial extension is given by \[Tr+1\text{ }=\text{ }nCr\text{ }an-r\text{ }br\] The overall term for binomial \[\left( 1+x \right)2n\] is \[Tr+1\text{ }=\text{...
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
The overall term \[Tr+1\] in the binomial extension is given by \[Tr+1\text{ }=\text{ }nCr\text{ }an-r\text{ }br\] Here the binomial is \[\left( 1+x \right)n\] with\[a\text{ }=\text{ }1\] ,...
In the expansion of (1 + a)m+n, prove that coefficients of am and an are equal
We realize that the overall term \[Tr+1\] in the binomial extension is given by \[Tr+1\text{ }=\text{ }nCr\text{ }an-r\text{ }br\] Here\[n=\text{ }m+n\] , \[a\text{ }=\text{ }1\] and...
Find the middle terms in the expansions of
Find the middle terms in the expansions of
Find the 13th term in the expansion of
Find the 4th term in the expansion of
(x – 2y)12. Solution: The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br Here a= x, n =12, r= 3 and b = -2y By substituting the values we get T4 = 12C3 x9 (-2y)3 =...
Write the general term in the expansion of (x2 – y x)12, x ≠ 0
The overall term \[Tr+1\] in the binomial development is given by \[Tr+1\text{ }=\text{ }n\text{ }C\text{ }r\text{ }an-r\text{ }br\] Here\[n\text{ }=\text{ }12\] , \[a=\text{ }x2\] and...
Write the general term in the expansion of (x2 – y)6
The overall term \[Tr+1\] in the binomial development is given by \[Tr+1\text{ }=\text{ }n\text{ }C\text{ }r\text{ }an-r\text{ }br\ldots \text{ }\ldots \text{ }..\text{ }\left( I \right)\]...
Find the coefficient of a5b7 in (a – 2b)12
The overall term \[Tr+1\] in the binomial extension is given by \[Tr+1\text{ }=\text{ }n\text{ }C\text{ }r\text{ }an-r\text{ }br\] Here \[a\text{ }=\text{ }a,\text{ }b\text{ }=\text{ }-\text{...
Find the coefficient of
x5 in (x + 3)8 Solution: The general term Tr+1 in the binomial expansion is given by Tr+1 = n C r an-r br Here x5 is the Tr+1 term so a= x, b = 3 and n =8 Tr+1 = 8Cr x8-r 3r…………… (i) For finding out...
Prove that:
according to ques,
Show that 9n+1 – 8n – 9 is divisible by 64, whenever n is a positive integer
To show that $$ \[9n+1\text{ }\text{ }8n\text{ }\text{ }9\] is divisible by\[64\] , it must be shown that\[9n+1\text{ }\text{ }8n\text{ }\text{ }9\text{ }=\text{ }64\text{ }k\] , where k is some...
Find (x + 1)6 + (x – 1)6. Hence or otherwise evaluate
Utilizing binomial hypothesis the articulations, $$ \[\left( x\text{ }+\text{ }1 \right)6\] and \[\left( x\text{ }\text{ }1 \right)6\] can be communicated as \[\left( x\text{ }+\text{ }1...
Find (a + b)4 – (a – b)4. Hence, evaluate
Utilizing binomial hypothesis the articulation \[\left( a\text{ }+\text{ }b \right)4\] and\[\left( a\text{ }\text{ }b \right)4\] , can be extended \[\left( a\text{ }+\text{ }b \right)4\text{...
Using Binomial Theorem, indicate which number is larger (1.1)10000 or 1000
By parting the given $$ \[1.1\] and afterward applying binomial hypothesis, the initial not many terms of \[\left( 1.1 \right)10000\] can be acquired as \[\left( 1.1 \right)10000\text{ }=\text{...
solve (99)5
Given \[\left( 99 \right)5\] \[99\] can be composed as the aggregate or distinction of two numbers then binomial hypothesis can be applied. The given inquiry can be composed as \[99\text{ }=\text{...
solve the following
(101)4 Given (101)4 \[101\] can be communicated as the total or distinction of two numbers and afterward binomial hypothesis can be applied. The given inquiry can be composed as (101)4 = (100 + 1)4...
7. Solve (102)5
Given $$ \[\left( 102 \right)5\] \[102\] can be communicated as the total or contrast of two numbers. now we use binomial theorem, The given inquiry can be composed as \[102\text{ }=\text{...
(96)3
Given \[\left( 96 \right)3\] \[96\] can be communicated as the aggregate or contrast of two numbers and afterward binomial hypothesis can be applied. The given inquiry can be composed as \[96\text{...
Expand each of the expressions in Exercises 1 to 5.
From binomial hypothesis, given condition can be extended as
Expand each of the expressions in Exercises 1 to 5.
From binomial hypothesis, given condition can be extended as
Expand each of the expressions in Exercises 1 to 5.
\[\mathbf{3}.\text{ }{{\left( \mathbf{2x}\text{ }\text{ }\mathbf{3} \right)}^{\mathbf{6}}}\] From binomial hypothesis, given condition can be extended as
Expand each of the expressions in Exercises 1 to 5.
From binomial hypothesis extension we can compose as
Expand each of the expressions in Exercises 1 to 5.
(1 – 2x)5 Solution: From binomial theorem expansion we can write as (1 – 2x)5 = 5Co (1)5 – 5C1 (1)4 (2x) + 5C2 (1)3 (2x)2 – 5C3 (1)2 (2x)3 + 5C4 (1)1 (2x)4 – 5C5 (2x)5 = 1 – 5 (2x) + 10 (4x)2 – 10...