ANSWER
![\begin{gathered} \text{The first thre}e\text{ terms of the binomial expression is given as} \\ 1+13x+78x^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/yl8hkodgwpj7krgzgl6vb80amxv70hwvan.png)
Explanation:
Given the following function
![(1+x)^(13)](https://img.qammunity.org/2023/formulas/mathematics/college/bozmdyniys5t8vwgihbnmr6bs9l4v6n8jx.png)
The above expression is a binomial expression
To find the first three terms, we need to apply the binomial theorem
The general formula for binomial theorem is given below as
![^nC_rx^{n\text{ - r}}y^r](https://img.qammunity.org/2023/formulas/mathematics/college/2n1qje1rwir6hueil11s603fol8ojmhmkn.png)
Let x = 1 and y = x
n = 13
r ranges from zero to 13
![\begin{gathered} ^(13)C_(0\cdot)1^{13\text{ - 0}}\cdot x^0+^(13)C^{}_1\cdot1^{13\text{ - 1}}\cdot x^1+^(13)C_2\cdot1^{13\text{ - 2}}x^2 \\ \text{ Recall that, the combination formula is given as} \\ ^nC_r\text{ = }\frac{n!}{(n\text{ - r)!r!}} \\ (13!)/((13-0)!0!)\cdot1^{13\text{ - 0 }}\cdot x^0\text{ + }\frac{13!}{(13\text{ - 1)!1!}}\cdot1^{13\text{ -1}}\cdot x^1\text{ + }\frac{13!}{(13\text{ - 2)!2!}}\cdot1^{13\text{ - 2}}\cdot x^2 \\ (13!)/(13!)\cdot1^{13-\text{ 0}}\cdot x^0\text{ + }(13!)/(12!)\cdot1^(12)\cdot x^1\text{ + }(13!)/(11!2!)\cdot1^{13\text{ - 2}}\cdot x^2 \\ 1\cdot\text{ 1 }\cdot\text{ 1 + 13 }\cdot\text{ 1 }\cdot\text{ x + }78\cdot\text{ 1 }\cdot x^2 \\ 1+13x+78x^2 \end{gathered}](https://img.qammunity.org/2023/formulas/mathematics/college/jiwh06ri3mwjofkfb5138lso51copvylxg.png)
Hence, the first three terms of the binomial expression are given below as
![1+13x+78x^2](https://img.qammunity.org/2023/formulas/mathematics/college/otp6w26zwq5ie04xqy6vhm3ongplnpucdb.png)