Use (x+3y^2)^14 to answer the following questions:

Question

How many terms are there?

Is there one middle term or two middle terms? Explain.

Which numerical term(s) is (are) the middle term(s)?

Find the middle term(s).

Find the last term.

Answer 1

1)

(x+3y²)^14 

Compare given expression with standard form (a+b)ⁿ .

a = x

b = 3y²

n = 14

So total terms present in the expansion =  n + 1 =  14 + 1 = 15.

15 terms included in given expansion.

Answer 2

2)

If n is even. So there is only one middle term.

Here n = 14 is even.

Total number of terms ( 15 ) is odd and n is even.

There exists only one middle term.

Answer 3

3)

Middle term order = (n/2)+1 = (14/2) + 1 = 7 + 1 = 8

8^th numerical term is the middle term.

Answer 4

4)

8^th term is Middle term.

General term T_r+1 = ⁿC_r x^n-r y^r

T₈ = T₇₊₁ = ¹⁴C₇ x^{14 - 7} y⁷.

T₈  = ¹⁴C₇ x⁷ y⁷.= 3432 x⁷ y⁷

Middle term is  ¹⁴C₇ x⁷ y⁷.(or) 3432 x⁷ y⁷

Answer 5

5)

Last term = bⁿ

Substitute : b = 3y²  and n = 14  (  from problem1 )

Last term = (3y²)¹⁴ = 3¹⁴ y^2*14 = 3¹⁴ y^2*14 = 3¹⁴ y²⁸

Last term is 3¹⁴ y²⁸.