Match List - I with List - II: LIST I LIST II (P) If the fourth term in the expansion of ( a x + x 1 ) n is 2 5 then a is (1) 2 (Q) ∑ p = 1 4 ∑ r = p 4 4 C r r C p = k then 13 k is (2)

Question

Match List - I with List - II:

LIST I	LIST II
(P) If the fourth term in the expansion of $(\frac{x}{a} + \frac{1}{x})^{n}$ is $\frac{5}{2}$ then a is	(1) 2
(Q) $\sum_{p = 1}^{4} \sum_{r = p}^{4}^{4} C_{r}^{r} C_{p} = k$ then $\frac{k}{13}$ is	(2) 4
(R) The coefficient of $x^{13}$ in $(1 - x)^{5} (1 + x + x^{2} + x^{3})^{4}$ &	(3) 5
(S) $\sum_{r = 0}^{4}^{4} C_{r} (r - 2)^{2} = k$ then $\frac{k}{2}$ is	(4) 8
	(5) 13

TestHub · Accepted Answer

( P ) n C 3 ​ ( a x ​ ) n − 3 ( x 1 ​ ) 3 = 2 5 ​ ​ ⇒ ​ n − 6 = 0 ⇒ a 3 = 8 ​ a = 2 (Q) ∑ p = 1 4 ​ ∑ r = p 4 ​ r ! ( 4 − r )! ( r − p )! p ! 4 ! ​ = ∑ p = 1 4 ​ 4 C p ​ ∑ r = p 4 ​ ( r − p )! ( 4 − r )! ( 4 − p )! ​ = ∑ p = 1 4 ​ 4 C p ​ 2 4 − p = 3 4 − 2 4 = 65 (R) Coefficient of x 13 in ( 1 − x ) ( 1 − x 4 ) 4 = − ( − 4 C 3 ​ ) = 4 (S) ∑ r = 0 4 ​ 4 C r ​ ( r − 2 ) 2 = 16

Mathematics - Binomial Theorem Question with Solution | TestHub

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