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...
(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}...
(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...
(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) \\...
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...
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...
(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}...
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}...
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...
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}={...
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}...
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. ${...
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)...
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...
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}...
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 $x^{15}$,...
(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...
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}...
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}...
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...
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{...
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
\[0.99\] 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{...
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{...
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{...
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...
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{...
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 \[n\text{ }=\text{ }m\]...
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 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\[a\text{ }=\text{ }1\] ,...
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...
(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 =...
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...
The overall term \[Tr+1\] in the binomial development is given by \[Tr+1\text{ }=\text{ }n\text{ }C\text{ }r\text{ }an-r\text{ }br\ldots \text{ }\ldots \text{ }..\text{ }\left( I \right)\]...
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{...
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...
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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...
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...
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{...
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{...
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{...
(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...
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{...
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{...
From binomial hypothesis, given condition can be extended as
From binomial hypothesis, given condition can be extended as
\[\mathbf{3}.\text{ }{{\left( \mathbf{2x}\text{ }\text{ }\mathbf{3} \right)}^{\mathbf{6}}}\] From binomial hypothesis, given condition can be extended as
From binomial hypothesis extension we can compose as
(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...