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1

### WB JEE 2022

MCQ (Single Correct Answer)
English
Bengali

The number of zeros at the end of $$\left| \!{\underline {\, {100} \,}} \right.$$ is

A
21
B
22
C
23
D
24

$$\left| \!{\underline {\, {100} \,}} \right.$$ -এর শেষে শূণ্যের সংখ্যা হবে

A
21
B
22
C
23
D
24
2

### WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
The coefficient of a3b4c5 in the expansion of (bc + ca + ab)6 is
A
$${{12!} \over {3!4!5!}}$$
B
$${{6!} \over {3!}}$$
C
33
D
$$3\,.\,\left( {{{6!} \over {3!3!}}} \right)$$

## Explanation

Given, (bc + ca + ab)6

General term = $${{6!} \over {p!q!r!}}{(bc)^p}{(ca)^q}{(ab)^r}$$

$$= {{6!} \over {p!q!r!}}{(a)^{q + r}}{(b)^{p + r}}{(c)^{p + q}}$$

Coefficient of a3b4c5 in the expansion (bc + ca + ab)6

$$\therefore$$ q + r = 3, p + r = 4, p + q = 5

= 2(p + q + r) = 12

= p + q + r = 6

$$\therefore$$ p = 3, q = 2, r = 1

$$\therefore$$ Coefficient of a3b4c5 is $${{6!} \over {3!2!1!}} = 3\left( {{{6!} \over {3!3!}}} \right)$$
(bc + ca + ab)6 এর বিস্তৃতিতে a3b4c5 -এর সহগ হল
A
$${{12!} \over {3!4!5!}}$$
B
$${{6!} \over {3!}}$$
C
33
D
$$3\,.\,\left( {{{6!} \over {3!3!}}} \right)$$

## Explanation

দেওয়া, (bc + ca + ab)6

সাধারন শর্তাবলী = $${{6!} \over {p!q!r!}}{(bc)^p}{(ca)^q}{(ab)^r}$$

$$= {{6!} \over {p!q!r!}}{(a)^{q + r}}{(b)^{p + r}}{(c)^{p + q}}$$

সম্প্রসারণে a3b4c5 এর সহগ (bc + ca + ab)6

$$\therefore$$ q + r = 3, p + r = 4, p + q = 5

= 2(p + q + r) = 12

= p + q + r = 6

$$\therefore$$ p = 3, q = 2, r = 1

$$\therefore$$ a3b4c5 এর সহগ $${{6!} \over {3!2!1!}} = 3\left( {{{6!} \over {3!3!}}} \right)$$
3

### WB JEE 2021

MCQ (Single Correct Answer)
English
Bengali
For x$$\in$$R, x $$\ne$$ $$-$$1, if $${(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}} + ..... + {x^{2016}} = \sum\limits_{i = 0}^{2016} {{a_i}\,.\,{x^i}}$$, then a17 is equal to
A
$${{2016!} \over {17!1999!}}$$
B
$${{2016!} \over {16!}}$$
C
$${{2017!} \over {2000!}}$$
D
$${{2017!} \over {17!2000!}}$$

## Explanation

For x$$\in$$R, x $$\ne$$ $$-$$1

$${(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}} + ..... + {x^{2016}} = \sum\limits_{i = 0}^{2016} {{a_i}\,.\,{x^i}}$$

Here, coefficient of x17

$$= {}^{2016}{C_{17}} + {}^{2015}{C_{16}} + {}^{2014}{C_{15}} + .... + {}^{1999}{C_0}$$

$$= {}^{2016}{C_{1999}} + {}^{2015}{C_{1999}} + {}^{2014}{C_{1999}} + ... + {}^{1999}{C_{1999}}$$

($$\because$$ $${}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}$$)

$$= {}^{2017}{C_{2000}} = {{(2017)!} \over {(2000)!\,(2017 - 2000)!}}$$

$$= {{2017!} \over {(2000)!\,(17)!}}$$
$${(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}} + ..... + {x^{2016}} = \sum\limits_{i = 0}^{2016} {{a_i}\,.\,{x^i}}$$, সকল x$$\in$$R, x $$\ne$$ $$-$$1 হলে a17 হবে
A
$${{2016!} \over {17!1999!}}$$
B
$${{2016!} \over {16!}}$$
C
$${{2017!} \over {2000!}}$$
D
$${{2017!} \over {17!2000!}}$$

## Explanation

x$$\in$$R, x $$\ne$$ $$-$$1এর জন্য

$${(1 + x)^{2016}} + x{(1 + x)^{2015}} + {x^2}{(1 + x)^{2014}} + ..... + {x^{2016}} = \sum\limits_{i = 0}^{2016} {{a_i}\,.\,{x^i}}$$

এখানে, x17 এর সহগ

$$= {}^{2016}{C_{17}} + {}^{2015}{C_{16}} + {}^{2014}{C_{15}} + .... + {}^{1999}{C_0}$$

$$= {}^{2016}{C_{1999}} + {}^{2015}{C_{1999}} + {}^{2014}{C_{1999}} + ... + {}^{1999}{C_{1999}}$$ ($$\because$$ $${}^n{C_r} + {}^n{C_{r - 1}} = {}^{n + 1}{C_r}$$)

$$= {}^{2017}{C_{2000}} = {{(2017)!} \over {(2000)!\,(2017 - 2000)!}}$$

$$= {{2017!} \over {(2000)!\,(17)!}}$$
4

### WB JEE 2020

MCQ (Single Correct Answer)
English
Bengali
If c0, c1, c2, ......, c15 are the binomial coefficients in the expansion

of (1 + x)15, then the value of $${{{c_1}} \over {{c_0}}} + 2{{{c_2}} \over {{c_1}}} + 3{{{c_3}} \over {{c_2}}} + ... + 15{{{c_{15}}} \over {{c_{14}}}}$$ is
A
1240
B
120
C
124
D
140

## Explanation

$${(1 + x)^{15}} = {c_0} + {c_1}x + {c_0}{x^2} + ... + {c_{15}}{x^{15}}$$

Now, $${{{c_1}} \over {{c_0}}} + 2{{{c_2}} \over {{c_1}}} + 3{{{c_3}} \over {{c_2}}} + ... + 15{{{c_{15}}} \over {{c_{14}}}}$$

$${{15 + 1 - 1} \over 1} + {{2(15 + 1 - 2)} \over 2} + {{3(15 + 1 - 3)} \over 3} + ... + {{15(15 + 1 - 15)} \over {15}}$$
$$\left[ {{{{}^n{C_r}} \over {{}^n{C_{r - 1}}}} = {{n + 1 - r} \over r}\,Here,\,n = 15} \right]$$

$$\therefore$$ 15 + 14 + 13 + ... + 1 = $${{15 \times 16} \over 2} = 120$$

$${(1 + x)^{15}}$$ এর বিস্তৃতিতে $${c_0},\,{c_1},\,{c_2},\,.....\,{c_{15}}$$ দ্বিঘাত সহগ হলে $${{{c_1}} \over {{c_0}}} + 2{{{c_2}} \over {{c_1}}} + 3{{{c_3}} \over {{c_2}}} + .... + 15{{{c_{15}}} \over {{c_{14}}}}$$ এর মান হবে

A
1240
B
120
C
124
D
140

## Explanation

$${t_1} = r.{{15{C_r}} \over {15{C_{r - 1}}}} = {{r.\left| \!{\underline {\, {15} \,}} \right. } \over {\left| \!{\underline {\, r \,}} \right. \left| \!{\underline {\, {15 - r} \,}} \right. }}.{{\left| \!{\underline {\, {r - 1} \,}} \right. \left| \!{\underline {\, {15 - r + 1} \,}} \right. } \over {\left| \!{\underline {\, {15} \,}} \right. }}$$

$$= 16 = r$$

$$\sum\limits_{r = 1}^{15} {{t_1} = 16 \times 15 - (1 + 2 + 3 + .... + 15)}$$

$$= 240 - 120 = 120$$

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