If n is an integer greater than 1 , show that: a - C 1 n ( a - 1 ) + C 2 n ( a - 2 ) - … . . + ( - 1 ) n ( a - n ) = 0 .

Given, a - C 1 n ( a - 1 ) + C 2 n ( a - 2 ) - … . . + ( - 1 ) n ( a - n ) = 0 We know that, C 0 n - C 1 n + C 2 n - . . . . . - 1 n · C n n = 0 & C r n = n r · C r - 1 n - 1 . ⇒ ∑ r = 0 n - 1 r C r n a - r = a ∑ r = 0 n - 1 r · C r n - ∑ r = 0 n - 1 r · C r n · r = a C 0 n - C 1 n + C 2 n - . . . . . - 1 n · C n n - ∑ r = 1 n - 1 r · n r × r · C r - 1 n - 1 = a 0 + n C 0 n - 1 - C 1 n - 1 + C 2 n - 1 - . . . . . . . - 1 n · C n n - 1 = 0 + n 0 = 0

If n is an integer greater than 1 , show that: a - C 1 n ( a - 1 ) + C 2 n ( a - 2 ) - … . . + ( - 1 ) n ( a - n ) = 0 .

Given, a - C 1 n ( a - 1 ) + C 2 n ( a - 2 ) - … . . + ( - 1 ) n ( a - n ) = 0 We know that, C 0 n - C 1 n + C 2 n - . . . . . - 1 n · C n n = 0 & C r n = n r · C r - 1 n - 1 . ⇒ ∑ r = 0 n - 1 r C r n a - r = a ∑ r = 0 n - 1 r · C r n - ∑ r = 0 n - 1 r · C r n · r = a C 0 n - C 1 n + C 2 n - . . . . . - 1 n · C n n - ∑ r = 1 n - 1 r · n r × r · C r - 1 n - 1 = a 0 + n C 0 n - 1 - C 1 n - 1 + C 2 n - 1 - . . . . . . . - 1 n · C n n - 1 = 0 + n 0 = 0

If ( 1 + x ) n = p 0 + p 1 x + p 2 x 2 + p 3 x 3 + … … . , then prove that: a p 0 - p 2 + p 4 - … … . = 2 n 2 cos n π 4 b p 1 - p 3 + p 5 - … … . = 2 n 2 sin n π 4

Given binomial expansion, ( 1 + x ) n = p 0 + p 1 x + p 2 x 2 + p 3 x 3 + … … . , Replace, x → ι ⇒ ( 1 + ι ) n = p 0 + p 1 ι + p 2 ι 2 + p 3 ι 3 + … … . , = p 0 + p 1 ι - p 2 - p 3 ι + … … . … 1 Replace, x → - ι ⇒ ( 1 - ι ) n = p 0 - p 1 ι + p 2 ι 2 - p 3 ι 3 + … … . , = p 0 - p 1 ι - p 2 + p 3 ι + … … . … 2 a Addition of equations 1 and 2 we get, 1 + ι n + 1 - ι n = 2 p 0 - p 2 + p 4 - . . . . . . . ⇒ p 0 - p 2 + p 4 - . . . . . . . = 1 + ι n + 1 - ι 2 n = 2 e i π 4 n + 2 e - i π 4 n 2 = 2 n 2 · e ι n π 4 + 2 n 2 · e - ι n π 4 2 = 2 n 2 cos n π 4 + ι sin n π 4 + cos n π 4 - ι sin n π 4 2 = 2 n 2 2 cos n π 4 2 = 2 n 2 · cos n π 4 b Subtraction of equation 2 from equation 1 1 + ι n - 1 - ι n = 2 ι p 1 - p 3 + p 5 - . . . . . . . ⇒ p 1 - p 3 + p 5 - . . . . . . . = 1 + ι n - 1 - ι 2 ι n = 2 e i π 4 n - 2 e - i π 4 n 2 ι = 2 n 2 · e ι n π 4 - 2 n 2 · e - ι n π 4 2 ι = 2 n 2 cos n π 4 + ι sin n π 4 - cos n π 4 + ι sin n π 4 2 ι = 2 n 2 2 ι sin n π 4 2 ι = 2 n 2 sin n π 4

Prove that if ' p ' is a prime number greater than 2 , then ( 2 + 5 ) p - 2 p + 1 is divisible by p , where [ . ] denotes the greatest integer function.

Given, p is a prime number greater than 2 , We know that all prime numbers are odd numbers except 2 , so p will be an odd number. ⇒ 5 + 2 p = C 0 p 2 p 5 0 + C 1 p 2 p - 1 5 1 + C 2 p 2 p - 2 5 2 + . . . . . + C p p 2 0 5 p . . . . . 1 ⇒ 5 - 2 p = - C 0 p 2 p 5 0 + C 1 p 2 p - 1 5 1 - C 2 p 2 p - 2 5 2 + . . . . . + C p p 2 0 5 p . . . . . . . 2 By subtracting equation 2 from equation 1 we get, ⇒ 5 + 2 p - 5 - 2 p = 2 C 0 p 2 p 5 0 + C 2 p 2 p - 2 5 2 + . . . . . + C p - 1 p 2 1 5 p . . . . . 1 Now, 5 = 2 . 236 ⇒ 5 - 2 = 0 . 236 , this value and any power of this always lies between 0 & 1 , so the greatest integer of that number will be 0 . ⇒ 5 + 2 p = 2 p + 1 + 2 C 2 p 2 p - 2 5 2 + . . . . . + C p - 1 p 2 1 5 p . . . . . 1 ⇒ 5 + 2 p - 2 p + 1 = 2 C 2 p 2 p - 2 5 2 + . . . . . + C p - 1 p 2 1 5 p . . . . . 1 ⇒ 5 + 2 p - 2 p + 1 = 2 × 5 C 2 p 2 p - 2 + . . . . . + C p - 1 p 2 1 5 p - 2 . . . . . 1 According to question p should be odd prime number and 5 is also odd prime number, so 5 + 2 p - 2 p + 1 will be divisible by p .

If ∑ r = 0 n - 1 r · C r n 1 2 r + 3 r 2 2 r + 7 r 2 3 r + … … . to m terms ] = k 1 - 1 2 m n , then find the value of k .

Given, ∑ r = 0 n - 1 r · n C r 1 2 r + 3 r 2 2 r + 7 r 2 3 r + … … . . to m terms ] ⇒ ∑ r = 0 n - 1 r · n C r 1 2 r + 3 4 r + 7 8 r + … … to m terms ] ⇒ ∑ r = 0 n - 1 r · n C r 1 2 r + ∑ r = 0 n ( - 1 ) r · n C r 3 4 r + ∑ r = 0 n ( - 1 ) r · n C r 7 8 r + … … . . to m terms By using, ∑ r = 0 n - 1 r · C r n x r = 1 - x n We get, 1 - 1 2 n + 1 - 3 4 n + 1 - 7 8 n + … . to m terms = 1 2 n + 1 4 n + 1 8 n + … . to m terms = 1 2 n + 1 2 2 n + 1 2 3 n + … . . to m terms = 1 2 n 1 2 n m - 1 1 2 n - 1 = 1 2 n 1 - 2 m n 1 - 2 n 2 n 2 m n = 1 - 2 m n 1 - 2 n 1 2 m n = 1 1 - 2 n 1 2 m n - 2 m n 1 - 2 n 1 2 m n = 1 1 - 2 n 1 - 1 2 m n = k 1 - 1 2 m n ⇒ k = 1 2 n - 1 .

Given s n = 1 + q + q 2 + … … + q n & S n = 1 + q + 1 2 + q + 1 2 2 + … . + q + 1 2 n , q ≠ 1 , prove that n + 1 C 1 + n + 1 C 2 · s 1 + n + 1 C 3 · s 2 + … + n + 1 C n + 1 · s n = 2 n · S n

Given, S n = 1 + q + q 2 + … + q n and S n = 1 + q + 1 2 + q + 1 2 2 + … + q + 1 2 n , q ≠ 1 Now, S n = 1 + q + q 2 + … + q n ⇒ It is clear above expression is sum of n + 1 terms of GP whose first term is 1 and common ratio is q . So, S n = 1 - q n + 1 1 - q … i (sum of n terms of GP whose first term is a and common ratio is r , A n = a 1 - r n 1 - r , r ≠ 1 ) And S n = 1 + q + 1 2 + q + 1 2 2 + … + q + 1 2 n , q ≠ 1 Above expression is also sum of n + 1 terms of GP whose first term is 1 and common ratio is q + 1 2 . So, S n = 1 - q + 1 2 n + 1 1 - q + 1 2 ⇒ S n = 2 n + 1 - q + 1 n + 1 2 n 1 - q Now we have to prove C 1 n + 1 + n + 1 C 2 s 1 + n + 1 C 3 s 2 + … + n + 1 C n + 1 s n = 2 n S n Let's take LHS So, LHS = C 1 n + 1 + n + 1 C 2 s 1 + n + 1 C 3 s 2 + … + n + 1 C n + 1 s n LHS = ∑ p = 1 n + 1 C p n + 1 s p - 1 LHS = ∑ p = 1 n + 1 C p n + 1 1 - q p 1 - q {from equation i } LHS = 1 1 - q ∑ p = 1 n + 1 C p n + 1 - ∑ p = 1 n + 1 C p n + 1 q p As we know from binomial theorem, 1 + x n + 1 = C 0 n + 1 + ∑ r = 1 n + 1 C r n + 1 x r So, ∑ p = 1 n + 1 C p n + 1 = 1 + 1 n + 1 - 1 and ∑ p = 1 n + 1 C p n + 1 q p = 1 + q n + 1 - 1 So, LHS = 1 1 - q 2 n + 1 - 1 - 1 + q n + 1 - 1 LHS = 2 n + 1 - 1 + q n + 1 1 - q LHS = 2 n · 2 n + 1 - 1 + q n + 1 2 n 1 - q LHS = 2 n · S n LHS = RHS Hence proved.

If 1 + x 15 = C 0 + C 1 · x + C 2 · x 2 + … . + C 15 · x 15 , then find the value of C 2 + 2 C 3 + 3 C 4 + … . + 14 C 15

Given, 1 + x 15 = C 0 + C 1 · x + C 2 · x 2 + . . . . . . + C 15 · x 15 ⇒ 1 + x 15 x = C 0 x + C 1 + C 2 · x + C 3 · x 2 + C 4 · x 3 + . . . . . . + C 15 · x 14 Differentiate both sides with respect to x ' ' ⇒ x · 15 1 + x 14 - 1 + x 15 x 2 = - C 0 x 2 + 0 + C 2 + 2 C 3 · x + 3 C 4 · x 2 + . . . . . + 14 C 15 · x 13 Put x = 1 on both sides, we get 15 . 2 14 - 2 15 1 = - 1 1 + C 2 + 2 C 3 + 3 C 4 + . . . . . + 14 C 15 ⇒ C 2 + 2 C 3 + 3 C 4 + . . . . . . + 14 C 15 = 2 14 15 - 2 + 1 = 2 14 . 13 + 1 = 212993

Prove that 1 2 C 1 n - 2 3 C 2 n + 3 4 C 3 n - 4 5 C 4 n + … . . + ( - 1 ) n + 1 n n + 1 · C n n = 1 n + 1

Let, 1 - x n = C 0 n - C 1 n x + C 2 n x 2 - C 3 n x 3 + . . . . . + C n n - x n Integrate both sides with respect to ' x ' 1 - x n + 1 - 1 n + 1 = C 0 n x - C 1 n x 2 2 + C 2 n x 3 3 - C 3 n x 4 4 + . . . . . + C n n - 1 n x n + 1 n + 1 ⇒ - 1 - x n + 1 x n + 1 = C 0 n - C 1 n x 2 + C 2 n x 2 3 - C 3 n x 3 4 + . . . . . + C n n - 1 n x n n + 1 Differentiate both sides with respect to ' x ' - 1 n + 1 x · n + 1 1 - x n - 1 - 1 - x n + 1 x 2 = 0 - C 1 n · 1 2 + C 2 n · 2 x 3 - C 3 n · 3 x 2 4 + . . . . . + C n n - 1 n n x n - 1 n + 1 Put x = 1 both sides, we get ⇒ 1 2 C 1 n - 2 3 C 2 n + 3 4 C 3 n - 4 5 C 4 n + . . . . . + - 1 n + 1 n + 1 = 1 n + 1

Prove that ∑ r = 0 n r 2 C r n p r q n - r = n p q + n 2 p 2 , if p + q = 1 .

Let ∑ r = 0 n r 2 C r n p r q n - r = ∑ r = 0 n r + r r - 1 C r n p r q n - r = ∑ r = 0 n r C r n p r q n - r + ∑ r = 0 n r r - 1 C r n p r q n - r = ∑ r = 1 n n p C r - 1 n - 1 p r - 1 q n - r + ∑ r = 2 n n n - 1 p 2 C r - 2 n - 2 p r - 2 q n - r = n p ∑ r - 1 n C r - 1 n - 1 p r - 1 q n - 1 - r - 1 + n n - 1 p 2 ∑ r = 2 n C r - 2 n - 2 p r - 2 q n - 2 - r - 2 = n p q + p n - 1 + n n - 1 p 2 q + p n - 2 Since it is given that p + q = 1 ⇒ ∑ r = 0 n r 2 C r n p r q n - r = n p 1 n - 1 + n 2 p 2 - n p 2 1 n - 2 = n p + n 2 p 2 - n p 2 = n p 1 - p + n 2 p 2 = n p q + n 2 p 2