Solve this:

Q9. The co-efficient of xk ( 0 k n ) in the expansion of E = 1 +(1+ x) + (1+x)2 +......+ (1 + x)is

         A   n + 1 C k + 1                                                                                                   B   n C k C   n + 1 C n - k - 1                                                                                                                     D   n + 1 C n - k - 1

Q10. The last term in the binomial expansion of  2 3 - 1 2   i s   1 3 . 9 3 log 3 8 . Then the 5th term from the beginning is 

            A   4 . 10 C 6                                                                                                                                   B   2 . 10 C 4   C   1 2 . 10 C 4                                                                                                                               C   10 C 6            

Kindly refer the similar  link for your query.
 10
https://www.meritnation.com/ask-answer/question/the-last-term-in-the-binomial-expansion-of-21-3-1-21-2-n/mathematical-induction-and-binomial-theorem/10198763
For remaining queries we request you to post them in separate threads to have rapid assistance from our experts.
Regards

  • 0
Ans) 10C6 

Process -->

We know that the last term of  ( ​√2 - 1/√​2 )n  is the first term of ​( ​ 1/√​2 - ​√2 )n

T1nC0  * ​ (1/​2)n
    ​  =  2(-n/2)

which according to the question 
is also 
=                    1       
                 (3 * 3√9) ^ log38

=    0.03141000804
... 2(-n/2) = 0.03141000804
= (-n/2) = log20.03141000804
          = -4.9926318775​
                      ​=   -5
... n = 10
So the 5th term from the beginning is (​( ​√2 - 1/√​2 )n)
T4+110C4 * ​​(√2)10-4 * ​(-1/√​2)4
        ​=  ​10C* 4 * (1/4)
= ​10C
Which can also be written as 10C6

 
  • 0
What are you looking for?