How many distinct term will be there in the expansion through binomial
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➡️Here is your answer through a Binomial Solution :-
The number of terms in (x+y)n(x+y)n, when expanded is n+1n+1. Now, if we look at (2x+3y−4z)n(2x+3y−4z)n and let w=2x+3yw=2x+3y, then you have (w−4z)n(w−4z)n, which has n+1n+1 terms.
Each of these terms is of the form C(2x+3y)a(−4z)n−aC(2x+3y)a(−4z)n−a, where CC is an appropriate constant. Since each of these terms has a different power on zz, when you expand the first part, there will be no cancellation.
Since aa can vary between 00 and nn, and we know how many terms (2x+3y)a(2x+3y)aexpands into, we get that the number of terms is
1+2+⋯+(n+1)=∑a=0na+1.1+2+⋯+(n+1)=∑a=0na+1.
This has a standard formula and is
(n+1)(n+2)2.
✌️ I THINK IT HELPED YOU ✌️
➡️Here is your answer through a Binomial Solution :-
The number of terms in (x+y)n(x+y)n, when expanded is n+1n+1. Now, if we look at (2x+3y−4z)n(2x+3y−4z)n and let w=2x+3yw=2x+3y, then you have (w−4z)n(w−4z)n, which has n+1n+1 terms.
Each of these terms is of the form C(2x+3y)a(−4z)n−aC(2x+3y)a(−4z)n−a, where CC is an appropriate constant. Since each of these terms has a different power on zz, when you expand the first part, there will be no cancellation.
Since aa can vary between 00 and nn, and we know how many terms (2x+3y)a(2x+3y)aexpands into, we get that the number of terms is
1+2+⋯+(n+1)=∑a=0na+1.1+2+⋯+(n+1)=∑a=0na+1.
This has a standard formula and is
(n+1)(n+2)2.
✌️ I THINK IT HELPED YOU ✌️
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