i have a homework with 10 question but im stuck with 3 i searched about them everywhere read other colleges lectures but i couldnt solved them finally i desired to ask hereQuestion-2:Prove that each positive integer can be written in form of 2 k ∗ q, where q is odd, and k is a non-negative integer.Hint: Use induction, and the fact that the product of two odd numbers is odd.Question-6: ( x + y ) n = ∑ k = 0 n C ( n , k ) ∗ x n − k ∗ y k = n 2 + n 2 Prove the above statement by using induction on n.Question-7:Let n 1 , n 2 , . . . , n t be positive integers. Show that if n 1 + n 2 + . . . + n t − t + 1 objects are placed into t boxes, then for some i, i = 1 , 2 , 3 , . . . , t , the ith box contains at least n i objects.

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i have a homework with 10 question but im stuck with 3 i searched about them everywhere read other colleges lectures but i couldnt solved them finally i desired to ask hereQuestion-2:Prove that each positive integer can be written in form of       2    k    ∗  q, where q is odd, and k is a non-negative integer.Hint: Use induction, and the fact that the product of two odd numbers is odd.Question-6:  (  x  +  y      )    n    =      ∑          k      =      0        n    C  (  n  ,  k  )  ∗      x          n      −      k        ∗      y    k    =                              n          2                +        n            2      Prove the above statement by using induction on n.Question-7:Let       n    1    ,      n    2    ,  .  .  .  ,      n    t   be positive integers. Show that if       n    1    +      n    2    +  .  .  .  +      n    t    −  t  +  1 objects are placed into t boxes, then for some i,   i  =  1  ,  2  ,  3  ,  .  .  .  ,  t , the ith box contains at least       n    i   objects.

Kylan Simon · Accepted Answer

(2) Certainly we can write   1  =      2    0    ⋅  1. Now suppose that   n  &amp;gt;  1, and every positive integer less than n can be written in the desired form. (This is your induction hypothesis.) Then either   n is odd, or n is even. If   n is odd, we can write   n  =      2    0    ⋅  n to express   n in the desired form. If   n is even, then   n  =  2  m for some positive integer   m  &amp;lt;  n. By the induction hypothesis there are a non-negative integer   k and an odd integer   q such that   m  =      2    k    ⋅  q; how can you use this to express   n in the desired form? (This is not quite the proof that the author of the hint had in mind; it’s actually a bit easier.)(6) As stated, this does not make sense. It’s true that  (  x  +  y      )    n    =      ∑          k      =      0        n                      (                    n      k                      )                  x          n      −      k            y    k      ;this is the binomial theorem. It is not true that this equals                               n          2                +        n            2      ; this is obvious just from the fact that                               n          2                +        n            2       doesn’t depend on   x and   y. What is true is that      ∑          k      =      0        n    k  =                    n        2            +      n        2      ;is this what you’re actually supposed to prove?

agrejas0hxpx · Accepted Answer

(7) Let       m    i   be the number of objects in the   i-th box, and suppose that       m    i    &amp;lt;      n    i   for   i  =  1  ,  …  ,  t. Then       m    i    ≤      n    i    −  1 for   i  =  1  ,  …  ,  t, so how big can       ∑          i      =      1        t        m    i   be?

i have a homework with 10 question but im stuck with 3 i searched about them everywhere read other c

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