How do you use Pascal’s triangle to expand ${{\left( 2a+1 \right)}^{5}}$ ?

Hint: The process of writing Pascal’s triangle is the all the rows of the triangle starts with 1 second term of nth row is sum of first and second term of n-1 throw , third term of the row is sum of second and third term of n-1th row and this goes on and the last term is equal to 1. So total n row present in nth row

Complete step by step answer:
We know nth row of Pascal’s triangle represent the coefficients of terms present in the expansion of ${{\left( 1+x \right)}^{n-1}}$
So n+1th row will represent the coefficients of terms present in the expansion od ${{\left( 1+x \right)}^{n}}$
That means the terms present in n+1th row are $^{n}{{C}_{0}}$ , $^{n}{{C}_{1}}$, $^{n}{{C}_{2}}$ , ……
So if we have to write the expansion of ${{\left( 2a+1 \right)}^{5}}$ using Pascal’s triangle we have look into the terms of row 6
So let’s draw the Pascal’s triangle
      
We can see the sixth row contains 1, 5, 10, 10, 5, 1
There are total 6 term present in the expansion of ${{\left( 2a+1 \right)}^{5}}$
1 is coefficient of ${{\left( 2a \right)}^{0}}$ , 5 is the coefficient of ${{\left( 2a \right)}^{1}}$ , 10 is the coefficient of ${{\left( 2a \right)}^{2}}$ and this goes on.
So the expansion of ${{\left( 2a+1 \right)}^{5}}$ is
$1\times {{\left( 2a \right)}^{0}}+5\times {{\left( 2a \right)}^{1}}+10\times {{\left( 2a \right)}^{2}}+10\times {{\left( 2a \right)}^{3}}+5\times {{\left( 2a \right)}^{4}}+1\times {{\left( 2a \right)}^{0}}$
$=1+10a+40{{a}^{2}}+80{{a}^{3}}+80{{a}^{4}}+32{{a}^{5}}$

Note:
Always keep in mind that the nth row of Pascal’s triangle represents the coefficient of terms present in the expansion of ${{\left( 1+x \right)}^{n-1}}$ so nth row contains the term $^{n-1}{{C}_{0}}$ , $^{n-1}{{C}_{1}}$ , $^{n-1}{{C}_{2}}$ ,
….., $^{n-1}{{C}_{n-1}}$ . So the nth row of Pascal’s triangle consists of total n terms.