Q: For nonnegative integers s and r , let s r = s ! r ! s - r ! if r ≤ s 0 if r > s . For positive integers m and n , let g m , n = ∑ p = 0 m + n f m ,

f m , n , p = ∑ i = 0 p m i n + i p p + n p - i = ∑ i = 0 p C i m · n + i ! p ! n + i - p ! · ( p + n ) ! p - i ! n + i ! = ∑ i = 0 p C i m · ( p + n ) ! p ! n ! · n ! n + i - p ! p - i ! = ∑ i = 0 p C i m . C p p + n · C p - i n C p p + n · ∑ i = 0 p C i m · C p - i n ∵ ∑ i = 0 p C i m · C p - i n = C p m + n ∴ f m , n , p = C p p + n . C p m + n ⇒ f m , n , p C p p + n = C p m + n . g m , n = ∑

Q: Suppose det ⁡ ∑ k = 0 n k ∑ k = 0 n C k n k 2 ∑ k = 0 n C k n k ∑ k = 0 n C k n 3 k = 0 , holds for some positive integer n . Then ∑ k = 0 n C

As given ∑ k = 0 n k ∑ k = 0 n C n k k 2 ∑ k = 0 n C n k k ∑ k = 0 n C n k 3 k = 0 a ∑ k = 0 n k = n n + 1 2 b ∑ k = 0 n C k n k 2 = ∑ k = 0 n k 2 - k + k n C k = ∑ k = 0 n k 2 - k C k n + ∑ k = 0 n k C k n = ∑ k = 0 n k k - 1 n k . n - 1 k - 1 C k - 2 n - 2 + ∑ k = 0 n k n k C k - 1 n - 1 = n n - 1 ∑ k = 0 n C k - 2 n - 2 + n ∑ k = 0 n C k -

Question 1

Let a_0, a_1, , a_ 23 be real numbers such that (1+ 2 5 x )^ 23 = _ i=0 ^ 23 a_i x^i for every real number x. let a_r be the largest among the numbers a_j for 0 j 23. The the value of r is ______

Accepted Answer

For x =1 (1+ 2 5 )^ 23 =a_0+a_1+a_2+ +a_ 23 for numerically greatest term n+1 1+ | a b | = 23+1 1+ 5 2 = 48 7 [ 48 7 ]=6= m (where [.] greatest integer function) so, T _7 is numerical greatest term Hence r=6

Question 2

Let a and b be two nonzero real numbers. If the coefficient of x 5  in the expansion of a x 2 + 70 27 b x 4  is equal to the coefficient of x - 5 in the expansion of a x - 1 b x 2 7 , then t

Accepted Answer

Given, The coefficient of  x 5  in  a x 2 + 70 27 b x 4  is equal to coefficient of  x - 5  in  a x - 1 b x 2 7 , Now finding the coefficient of  x 5  in  a x 2 + 70 27 b x 4  we get, T r + 1 = C r 4 a x 2 4 - r &#183; 70 27 b x r &#8658; T r + 1 = C r 4 a 4 - r &#183; 70 27 b r &#183; x 8 - 2 r - r So,  8 - 2 r - r = 5 &#8658; r = 1 So,  T 2 = C 1 4 a 3 &#183; 70 27 b &#183; x 5 Now finding the coefficient of  x - 5  in&

Question 3

For nonnegative integers s and r , let  s r = s ! r ! s - r ! &#160;if&#160; r &#8804; s 0 &#160;if&#160; r &#62; s . For positive integers m and n , let  g m , n = &#8721; p = 0 m + n f m ,

Accepted Answer

f m , n , p = &#8721; i = 0 p m i n + i p p + n p - i = &#8721; i = 0 p C i m &#183; n + i ! p ! n + i - p ! &#183; ( p + n ) ! p - i ! n + i ! = &#8721; i = 0 p C i m &#183; ( p + n ) ! p ! n ! &#183; n ! n + i - p ! p - i ! = &#8721; i = 0 p C i m . C p p + n &#183; C p - i n C p p + n &#183; &#8721; i = 0 p C i m &#183; C p - i n &#8757; &#8721; i = 0 p C i m &#183; C p - i n = C p m + n &#8756; f m , n , p = C p p + n . C p m + n &#8658; f m , n , p C p p + n = C p m + n . g m , n = &#8721;

Question 4

Suppose det &#8289; &#8721; k = 0 n k &#8721; k = 0 n C k &#160; n k 2 &#8721; k = 0 n C k &#160; n k &#8721; k = 0 n C k &#160; n 3 k = 0 , holds for some positive integer n . Then &#8721; k = 0 n C

Accepted Answer

As given &#8721; k = 0 n k &#8721; k = 0 n C &#160; n k k 2 &#8721; k = 0 n C &#160; n k k &#8721; k = 0 n C &#160; n k 3 k = 0 a &#160; &#8721; k = 0 n k = n n + 1 2 b &#160; &#8721; k = 0 n C k &#160; n k 2 = &#8721; k = 0 n k 2 - k + k n C k = &#8721; k = 0 n k 2 - k C k &#160; n + &#8721; k = 0 n k C k &#160; n = &#8721; k = 0 n k k - 1 n k . n - 1 k - 1 C k - 2 &#160; n - 2 + &#8721; k = 0 n k n k C k - 1 &#160; n - 1 = n n - 1 &#8721; k = 0 n C k - 2 &#160; n - 2 + n &#8721; k = 0 n C k -

Binomial Theorem — Previous Year Questions & Solutions