Question

Fill in the blanks.
(i) If $25^{15}$ is divided by $13$, then the remainder is P.
(ii) The coefficient of $x$ in the expansion of $(1-3x+7x^2)(1-x{)}^{16}$ is Q.
(iii) If $n$ is even and the middle term in the expansion of ${\left(x^2+\frac{1}{x}\right)}^n$ is $924x^6$, then n is equal to R.
(iv) The number of terms in the expansion of $[(2x+y^3{)}^4{]}^7$ is S.

	P	Q	R	S
(a)	14	-15	12	29
(b)	14	-15	10	18
(c)	12	-19	10	29
(d)	12	-19	12	29

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (D)

Let $25^{15}=(26-1{)}^{15}$
$={}^{15}{C}_{0}26^{15}-{}^{15}{C}_{1}26^{14}+......-{}^{15}{C}_{15}$
$={}^{15}{C}_{0}26^{15}-{}^{15}{C}_{1}26^{14}+......-1-13+13$
$={}^{15}{C}_{0}26^{15}-{}^{15}{C}_{1}26^{14}+.....-13+12$
$=13k+12$, where $k\in N$
Hence, when $25^{15}$ is divided by $13$, then remainder will be $12$.
(ii) Given expansion is $(1-3x+7x^2)(1-x{)}^{16}$
$=(1-3x+7x^2){(}^{16}{C}_0-{}^{16}{C}_1{x}^1+{}^{16}{C}_2{x}^2+....+{}^{16}{C}_{16}x{1}^6)$
$=(1-3x+7x^2)(1-16x+120x^2+.....+x^{16})$
$\therefore$ Coefficient of $x=-3-16=-19$
(iii) Since, $n$ is even, therefore ${\left(\frac{n}{2}+1\right)}^{th}$ term is the middle term
$\therefore T_{\frac{n}{2}+1}={}^n{C}_{n/2}{\left(x^2\right)}^{n/2}{\left(\frac{1}{x}\right)}^{n/2}=924x^6$ (given)
On comparing, we get $x^{n/2}=x^6$
$\Rightarrow n=12$
(iv) We have, $[(2x+y^3{)}^4{]}^7=(2x+y^3{)}^{28}$
$\therefore$ This expansion has $29$ terms.