Q7. Expand (5-3x)^{3} by Pascal's triangle
(A) 125+225x+135x^{2}+27x^{3}+\cdot\cdot\cdot
(C) 125-225x-135x^{2}+27x^{3}+\cdot\cdot\cdot
(B) 125+225x+135x^{2}-27x^{3}+\cdot\cdot\cdot
(D) 125-225x+i35x^{2}-27x^{3}+\cdot\cdot\cdot
To expand the expression (5-3x)^3 using Pascal's triangle, we need to find the coefficients of each term.
The first row of Pascal's triangle is always 1. The second row is 1 1. The third row is obtained by adding adjacent elements from the second row, so it is 1 2 1. Similarly, in the fourth row, we have 1 3 3 1, and so on.
Let's use a modified Pascal's triangle for our expansion:
1
1 1
1 2 1
1 3 3 1
...
To find the coefficients of the terms in the expansion of (5-3x)^3, we start with the highest power of x and work our way down.
The highest power of x in (5-3x)^3 is x^3. Using the first row of the modified Pascal's triangle, which is 1, our first term is 1 * 5^3 * (-3x)^0 = 125.
For the next term, which has x^2, we use the second row of the modified Pascal's triangle, which is 1 1. We have 1 * 5^2 * (-3x)^1 = 225x.
For the third term, which has x, we use the third row of the modified Pascal's triangle, which is 1 2 1. We have 2 * 5^1 * (-3x)^2 = 135x^2.
For the last term, which has no x, we use the fourth row of the modified Pascal's triangle, which is 1 3 3 1. We have 3 * 5^0 * (-3x)^3 = -27x^3.
Therefore, the expanded form of (5-3x)^3 by Pascal's triangle is 125 + 225x + 135x^2 - 27x^3.
So, the correct answer is (B) 125+225x+135x^2-27x^3+...