Binomial

Find the number of terms in the expansion of (x^2 + 1 + 1/x^2)^n ,n=N

10 Answers

62
Lokesh Verma ·

This one is not very difficult if you think of the number of solutions of

2a-2b.. where a+b+c=n and a,b,c ≥0

Do you get the hint?

1
Rocky Crazy ·

Nt getting u boss..

1
Jagaran Chowdhury ·

no. of terms in the expansion of(x+y+z)n

=(x+(y+z ))n
=xn*1+xn-1*(y+z)+xn-2*(y+z)2+......

=1 term +2 terms +3terms+.....(n+1) terms
=1/2(n+1)(n+2) terms

62
Lokesh Verma ·

no jagaran.. you have to be careful.. this will not work here..

1
Jagaran Chowdhury ·

@nishant sir, please explain ur hint ,sir and is the solution 1/2(n+1)(n+2) correct?

1
Jagaran Chowdhury ·

got it sir terms with similar powers of x club together reducing no. of terms.

62
Lokesh Verma ·

yes perfect..

now refer to my hint :)

1
Rocky Crazy ·

I m not getting sir please give a little more hint..

62
Lokesh Verma ·

Each term will be ....

(x2)a1b(x-2)c where a+b+c = n

1
Rocky Crazy ·

Ok..i got it..Thanks a lot sir..Thanks alot..

Your Answer

Close [X]