In the expansion of \({\left( {\sqrt {\rm{x}} + \frac{1}{{3{{\rm{x}}^2}}}} \right)^{10}}\) the value of constant term (independent of x) is
5
Finding the Constant Term in Binomial Expansion
The question asks for the value of the constant term, which is the term independent of \(x\), in the expansion of \({\left( {\sqrt {\rm{x}} + \frac{1}{{3{{\rm{x}}^2}}}} \right)^{10}}\).
We use the binomial theorem to find the general term of the expansion. The binomial theorem states that the expansion of \({\left( {a + b} \right)^n}\) is given by:
\[{\left( {a + b} \right)^n} = \sum_{r=0}^{n} \binom{n}{r} a^{n-r} b^r\]The general term, often denoted as \(T_{r+1}\), is given by:
\[T_{r+1} = \binom{n}{r} a^{n-r} b^r\]In our given expression, \({\left( {\sqrt {\rm{x}} + \frac{1}{{3{{\rm{x}}^2}}}} \right)^{10}}\):
- The first term is \(a = \sqrt{x} = x^{1/2}\).
- The second term is \(b = \frac{1}{{3{{\rm{x}}^2}}} = \frac{1}{3}x^{-2}\).
- The power of the expansion is \(n = 10\).
Substituting these values into the formula for the general term \(T_{r+1}\):
\[T_{r+1} = \binom{10}{r} \left(x^{1/2}\right)^{10-r} \left(\frac{1}{3}x^{-2}\right)^r\]Now, let's simplify the terms involving \(x\):
\[T_{r+1} = \binom{10}{r} x^{(1/2)(10-r)} \left(\frac{1}{3}\right)^r (x^{-2})^r\] \[T_{r+1} = \binom{10}{r} \left(\frac{1}{3}\right)^r x^{5 - r/2} x^{-2r}\]Combine the powers of \(x\) using the rule \(x^m x^n = x^{m+n}\):
\[T_{r+1} = \binom{10}{r} \left(\frac{1}{3}\right)^r x^{5 - r/2 - 2r}\] \[T_{r+1} = \binom{10}{r} \left(\frac{1}{3}\right)^r x^{5 - 5r/2}\]For the term to be independent of \(x\) (the constant term), the exponent of \(x\) must be zero. So, we set the power of \(x\) equal to 0 and solve for \(r\):
\[5 - \frac{5r}{2} = 0\] \[5 = \frac{5r}{2}\]Multiply both sides by 2:
\[10 = 5r\]Divide both sides by 5:
\[r = \frac{10}{5} = 2\]So, the constant term corresponds to \(r=2\). Now, we substitute \(r=2\) back into the expression for \(T_{r+1}\) to find the value of this term:
The constant term is \(T_{2+1} = T_3\).
\[T_3 = \binom{10}{2} \left(\frac{1}{3}\right)^2 x^{5 - 5(2)/2}\] \[T_3 = \binom{10}{2} \left(\frac{1}{9}\right) x^{5 - 5}\] \[T_3 = \binom{10}{2} \left(\frac{1}{9}\right) x^0\]Since \(x^0 = 1\), the constant term is \(\binom{10}{2} \times \frac{1}{9}\).
Now, we calculate the binomial coefficient \(\binom{10}{2}\):
\[\binom{10}{2} = \frac{10!}{2!(10-2)!} = \frac{10!}{2!8!} = \frac{10 \times 9}{2 \times 1} = 5 \times 9 = 45\]Now substitute this value back into the expression for the constant term:
\[\text{Constant Term} = 45 \times \frac{1}{9}\] \[\text{Constant Term} = \frac{45}{9} = 5\]Thus, the value of the constant term in the expansion is 5.
Revision Table: Key Concepts
| Concept | Description |
|---|---|
| Binomial Theorem | Formula for expanding \({\left( {a + b} \right)^n}\). |
| General Term \(T_{r+1}\) | The \((r+1)\)-th term in the expansion, given by \(\binom{n}{r} a^{n-r} b^r\). |
| Constant Term | A term in the expansion that does not contain the variable \(x\). Its exponent for \(x\) is 0. |
Additional Information on Binomial Expansion
When working with binomial expansions like \({\left( {a + b} \right)^n}\), understanding the general term \(T_{r+1}\) is crucial, especially for finding specific terms such as the constant term, terms with a specific power of \(x\), or the middle term(s).
- The value of \(r\) ranges from 0 to \(n\), where \(n\) is the power of the binomial.
- The sum of the powers of \(a\) and \(b\) in each term is always equal to \(n\), i.e., \((n-r) + r = n\).
- To find a term with a specific power of \(x\), set the exponent of \(x\) in the general term \(T_{r+1}\) equal to the desired power and solve for \(r\). If the value of \(r\) is a non-negative integer between 0 and \(n\) (inclusive), then such a term exists.
- The coefficients \(\binom{n}{r}\) are called binomial coefficients and can be found using Pascal's triangle or the formula \(\frac{n!}{r!(n-r)!}\).
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