Prove that ∑ k = 0 n C k n sin k x · cos n - k x = 2 n - 1 sin n x .

Keep the values of k and add for series, we get S = C 0 n · sin 0 x · cos n x + C 1 n · sin x · cos n - 1 x … … + C n n sinn x · cos 0 x … … 1 Reversing the series we get, S = C 0 n · sin n x · cos 0 x + C 1 n · sin n - 1 x · cos x + … … + C n n sin 0 x cos n x … … 2 Add 1 and 2 using sin A . cos B + cos A . sin B = sin A + B , we get 2 S = C 0 n + C 1 + n + … + C n n sin n x ⇒ S = 2 n - 1 sin n x

Prove that: ∑ r = 0 25 30 r 70 25 - r = C 25 100

L H S = ∑ r = 0 25 30 r 70 25 - r = ∑ r = 0 25 C r 30 · C 25 - r 70 = Coefficient of x 25 in the expansion of 1 + x 30 · 1 + x 70 ,i.e., 1 + x 100 . = C 25 100 = R H S .

Prove that: n r r k = n k n - k r - k

L H S = C r n · C k r = n ! r ! n - r ! · r ! k ! r - k ! = n ! k ! n - k ! · n - k ! r - k ! n - r ! = C k n · C r - k n - k = n k · n - k r - k = R H S

Prove that: ∑ r = 0 25 - 1 r 30 r 30 25 - r = 0

LHS = ∑ r = 0 25 - 1 r 30 r 30 25 - r = C 0 30 C 25 30 - C 1 30 C 24 30 + C 2 30 C 23 30 - . . . - C 25 30 · C 0 30 Now, 1 + x 30 = C 0 30 + C 1 30 x + C 2 30 x 2 + . . . + C 23 30 x 23 + C 24 30 x 24 + C 25 30 x 25 + . . . + C 30 30 x 30 . . . 1 1 - x 30 = C 0 30 - C 1 30 x + C 2 30 x 2 + . . . - C 23 30 x 23 + C 24 30 x 24 - C 25 30 x 25 + . . . + C 30 30 x 30 . . . 2 Multiplying equations (1) & (2) & comparing coefficient of x 25 on both sides, Coefficient of x 25 in 1 - x 2 30 = C 0 - 30 C 25 30 + C 1 30 C 24 30 - C 2 30 C 23 30 + . . . - C 24 30 C 1 30 + C 25 30 C 0 30 General term of 1 - x 2 30 = - 1 r C r 30 x 2 30 - r = - 1 r C r 30 x 60 - 2 r But 60 - 2 r ≠ 25 for any non-negative integer of r ∴ Coefficient of x 25 in 1 - x 2 30 = 0 ∴ C 0 30 C 25 30 - C 1 30 C 24 30 + C 2 30 C 23 30 + . . . + C 24 30 C 1 30 - C 25 30 C 0 30 = 0 = RHS Hence, proved.

Prove that: ∑ r = 0 n - 2 n - 1 r n r + 2 = 2 n - 1 n - 2

∑ r = 0 n - 2 n - 1 r n r + 2 = 2 n - 1 n - 2 L H S = ∑ r = 0 n - 2 C n - r - 1 n - 1 · C r + 2 n = Coefficient of x n + 1 in the expansion of 1 + x n - 1 1 + x n ,i.e., 1 + x 2 n - 1 = C n + 1 2 n - 1 = C n - 2 2 n - 1 = 2 n - 1 n - 2 = R H S ∵ C r n = C n - r n

Prove that C 0 n C 1 n + C 1 n C 2 n + . . . + C n - 1 n C n n = 2 n ! n + 1 ! n - 1 ! .

LHS = C 0 n C 1 n + C 1 n C 2 n + . . . + C n - 1 n C n n Now, 1 + x n = C 0 n + C 1 n x + C 2 n x 2 + . . . + C n n x n . . . 1 1 + 1 x n = C 0 n + C 1 n 1 x + C 2 n 1 x 2 + . . . + C n n 1 x n x + 1 n x n = C 0 n + C 1 n 1 x + C 2 n 1 x 2 + . . . + C n n 1 x n . . . 2 Multiply by x n on both sides, x + 1 n = C 0 n x n + C 1 n x n - 1 + C 2 n x n - 2 + . . . . + C n n . . . . 3 Multiply equations 1 & 3 , 1 + x n x + 1 n = C 0 n + C 1 n x + C 2 n x 2 + . . . + C n n x n C 0 n x n + C 1 n x n - 1 + C 2 n x n - 2 + . . + C n - 1 n x + C n n Comparing coefficient of x n - 1 on both sides, Coefficient of x n - 1 in 1 + x 2 n = C 0 n C 1 n + C 1 n C 2 n + C 2 n C 3 n . . . + C n - 1 n C n n ⇒ C 0 n C 1 n + C 1 n C 2 n + . . . + C n - 1 n C n n = C n - 1 2 n = 2 n ! n - 1 ! n + 1 ! = RHS Hence, proved.

Prove that C 0 n 2 + C 1 n 2 + C 2 n 2 + . . . . . + C n n 2 = 2 n ! n ! n ! .

LHS = C 0 n 2 + C 1 n 2 + C 2 n 2 + . . . . . + C n n 2 Now, 1 + x n = C 0 n + C 1 n x + C 2 n x 2 + . . . + C n n x n . . . i x + 1 n = C 0 n x n + C 1 n x n - 1 + . . . . + C n n . . . ii Multiplying equations i & i i and comparing coefficients of x n on both sides, we get, C 0 n 2 + C 1 n 2 + C 2 n 2 + . . . . . + C n n 2 = coefficient of x n in 1 + x 2 n = C n 2 n = 2 n ! n ! n ! = RHS Hence, proved.

E M B I B E

Beta Question Bank for Engineering: Mathematics>Chapter 8 - Binomial Theorem>EXERCISE-4

Embibe Experts Solutions for Chapter: Binomial Theorem, Exercise 4: EXERCISE-4

Author:Embibe Experts

Embibe Experts Mathematics Solutions for Exercise - Embibe Experts Solutions for Chapter: Binomial Theorem, Exercise 4: EXERCISE-4

Attempt the free practice questions on Chapter 8: Binomial Theorem, Exercise 4: EXERCISE-4 with hints and solutions to strengthen your understanding. Beta Question Bank for Engineering: Mathematics solutions are prepared by Experienced Embibe Experts.

Questions from Embibe Experts Solutions for Chapter: Binomial Theorem, Exercise 4: EXERCISE-4 with Hints & Solutions

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 8 - Binomial Theorem>EXERCISE-4>Q 3

Prove that $\sum_{k = 0}^{n} {}^{n}C_{k} \sin k x \cdot \cos (n - k) x = 2^{n - 1} \sin n x$ .

MEDIUM

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 8 - Binomial Theorem>EXERCISE-4>Q 4

Find the coefficient of:

(a) $x^{6}$ in the expansion of ${(a x^{2} + b x + c)}^{9}$

(b) $x^{2} y^{3} z^{4}$ in the expansion of ${(a x - b y + c z)}^{9}$ .

(c) $a^{2} b^{3} c^{4} d$ in the expansion of ${(a - b - c + d)}^{10}$ .

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 8 - Binomial Theorem>EXERCISE-4>Q 5

Prove that: $\sum_{r = 0}^{25} (\begin{matrix} 30 \\ r \end{matrix}) (\begin{matrix} 70 \\ 25 - r \end{matrix}) = {}^{100}C_{25}$

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 8 - Binomial Theorem>EXERCISE-4>Q 6

Prove that: $(\begin{matrix} n \\ r \end{matrix}) (\begin{matrix} r \\ k \end{matrix}) = (\begin{matrix} n \\ k \end{matrix}) (\begin{matrix} n - k \\ r - k \end{matrix})$

HARD

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 8 - Binomial Theorem>EXERCISE-4>Q 7

Prove that: $\sum_{r = 0}^{25} {(- 1)}^{r} (\begin{matrix} 30 \\ r \end{matrix}) (\begin{matrix} 30 \\ 25 - r \end{matrix}) = 0$

EASY

JEE Main/Advance

IMPORTANT

Beta Question Bank for Engineering: Mathematics>Chapter 8 - Binomial Theorem>EXERCISE-4>Q 8

Prove that: $\sum_{r = 0}^{n - 2} (\begin{matrix} n - 1 \\ r \end{matrix}) (\begin{matrix} n \\ r + 2 \end{matrix}) = (\begin{matrix} 2 n - 1 \\ n - 2 \end{matrix})$