Expert: Scott A Wilson - 10/21/2011

Question
QUESTION: Sir I am having problem in the questions in the image.Sir plzzzz give clear and explanatory solutions.If there are any formulas or shortcuts of these questions,plzzz tell.

ANSWER: 1. The number of terms would depend on how many powers of x there would be.
Since we have x to the 0, 1, 2, and 3 all taken to the 6, all possibilities would exist between 0 and 18, giving 19 terms.  By the way, 1 + 3x + 3x² + x³ = (1+x)³, so this is (1+x)^18.
It would be x^18 + 18x^17 + 153x^16 + 816x^15 + 3,060x^14 + 8,568x^13 + 18,564x^12 + 31,824x^11 + 43,758x^10 + 48,620x^9 + 4,3758x^8 + 31,824x^7 + 18,564x^6 + 8,568x^5 + 3,060x^4 + 816x^3 + 153x^2 + 18x + 1.  This is interesting since the 3rd term from the start and from the end is 153, and 1³ + 5³ + 3³ = 1 + 125 + 27 = 153.

2. In (a+b)^2, the number of terms in (a^n)(b^(2-n) is C(2,n) where C(x,y) is x choose y.
When n = 0, it is C(2,0), which is 1; when n = 1, it is C(2,1), which is 2; when n = 2, it is C(2,2), which is 1.  Thus, the terms are 1 2 1 on a² + 2ab + b².
If we take (a+b)³, we get C(3,0), C(3,1), C(3,2), C(3,3) as the numbers.
This is 1 3 3 0 for a³ + 3a²b + 3ab² + b³.

For (x^3)(y^4)z in (1+x+y-z)^9, it would be C(9;3,4,1,1), which is 9!/(3!4!1!1!).
That is 362,880/(6*24*1*1) = 2,520.  Since there is only 1 z involved, it is -2,520.

Another note on the side: 3!4! = (3*4)², which is not normal since x!y! is generally not (xy)².

3. I have seen that to the 1st, the sum of the terms is -1, and
to the 2nd, 1 2 -5 -6 9 as the terms and the sum is 1, and that is where it ends.
When the term is taken to the 3rd, 4th, 5th, 6th, or 7th, the sum of the terms is 0.
I would expect this to carry on until 2163, so I would guess it to be 0 as well.
This is because if the numbers are taken by any constant and added up, the sum is still the same.
Before the final addition is done, the sum of the previous coefficients times 1 is the same, by x is the same, and -3x² is -3 times the same, which is still 0, so when added together, the result is still 0.

4. Since there are 3 terms to start with, the first and the last involving x^2, that adds two terms each time the power is increased by 1.  This means to the nth, there are 2n+1 terms.
Since only the center terms is not dependent on x, there are 2n+1-1 = 2n terms dependent on x.


---------- FOLLOW-UP ----------

QUESTION: I saw these questions in the topic multinomial expansion and they have formulas.plzzz tell me those pre-made formulas as I am preparing for competitive exams.

Answer
For question 1, the number of terms can be computed by knowing how many terms there are to start with and then looking at how many more would be generated when squaring it.  The number added when squaring the expression is the number that are added for each power over 1.

For question 2, the thing to know is that for any set of powers n1, n2, .,. nk,
the answer is C(n;n1, n2, .. nk) where n1 + n2 + ... + nk = n.
That is n!/(n1!n2!..nk!).

For question 3, I worked it out to the first few powers and saw that the sum of the coefficients went to 0.  This would imply that all future operations would have the sum of the coefficients be 0.

For question 4, I multiplied out the next two terms and saw that each multiplication added two terms, there was an extra term at the start, but that this term wasn't counted since it was a constant.

The basic way to do these problems is to see how many terms are added with each power
and note the effect of factorials when counting the number.