Find the first three terms in ascending power of 'x' of the binomial expression of (3+bx)^5 where b is a non zero constant.
3^5 + ***** 3^4 (bx) + ****** 3^3 (bx)^2 + ........
now get binomial coefficients for fifth power from Pascal's triangle or formula
1 5 10 10 5 1
so
3^5 + 5*3^4 b x + 10 * 3^3 b^2 x^2
243 + 405 b x + 270 b^2 x^2 ......
To find the first three terms in ascending power of 'x' of the binomial expression (3+bx)^5, we can use the Binomial Theorem. The Binomial Theorem states that for any positive integer n, the expansion of (a+b)^n can be written as:
(a+b)^n = C(n,0)*a^n*b^0 + C(n,1)*a^(n-1)*b^1 + C(n,2)*a^(n-2)*b^2 + ... + C(n,n)*a^0*b^n
Here, C(n,r) represents the binomial coefficient, which is calculated as C(n,r) = n! / (r!(n-r)!), where n! represents the factorial of n.
In our case, the binomial expression is (3+bx)^5, where b is a non-zero constant. Therefore, a = 3 and b = bx.
Let's now find the first three terms:
Term 1 (x^0): C(5,0)*3^5*(bx)^0 = 1*3^5*1 = 243
Term 2 (x^1): C(5,1)*3^4*(bx)^1 = 5*3^4*b = 405b
Term 3 (x^2): C(5,2)*3^3*(bx)^2 = 10*3^3*b^2*x^2 = 270b^2*x^2
Therefore, the first three terms in ascending power of 'x' of the binomial expression (3+bx)^5 are:
Term 1: 243
Term 2: 405b
Term 3: 270b^2*x^2
To find the first three terms in ascending power of 'x' of the binomial expression (3+bx)^5, we can expand the expression using the binomial theorem.
The binomial theorem states that for any non-negative integer n, we can expand the binomial expression (a + b)^n using the following formula:
(a + b)^n = C(n,0) * a^n * b^0 + C(n,1) * a^(n-1) * b^1 + C(n,2) * a^(n-2) * b^2 + ... + C(n,n-1) * a^1 * b^(n-1) + C(n,n) * a^0 * b^n
Where C(n, k) represents the binomial coefficient, which is calculated as: C(n, k) = n! / (k! * (n-k)!)
In this case, we are given the binomial expression (3 + bx)^5, where b is a non-zero constant. Applying the binomial theorem, we can expand the expression as:
(3 + bx)^5 = C(5,0) * 3^5 * (bx)^0 + C(5,1) * 3^4 * (bx)^1 + C(5,2) * 3^3 * (bx)^2 + C(5,3) * 3^2 * (bx)^3 + C(5,4) * 3^1 * (bx)^4 + C(5,5) * 3^0 * (bx)^5
Simplifying this expression will give us the binomial expansion with terms in powers of x. Let's break it down:
Term 1: C(5,0) * 3^5 * (bx)^0 = 1 * 3^5 * 1^0 = 3^5 = 243
Term 2: C(5,1) * 3^4 * (bx)^1 = 5 * 3^4 * b * x
Term 3: C(5,2) * 3^3 * (bx)^2 = 10 * 3^3 * (b^2) * (x^2)
Therefore, the first three terms in ascending power of 'x' of the binomial expression (3+bx)^5 are:
1) 243
2) 5 * 3^4 * b * x
3) 10 * 3^3 * (b^2) * (x^2)