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1

### JEE Main 2021 (Online) 27th August Morning Shift

$$\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}}$$ is equal to :
A
$${}^{40}{C_{21}}$$
B
$${}^{40}{C_{19}}$$
C
$${}^{40}{C_{20}}$$
D
$${}^{41}{C_{20}}$$

## Explanation

$$\sum\limits_{k = 0}^{20} {{{\left( {{}^{20}{C_k}} \right)}^2}}$$

= $${\left( {{}^{20}{C_0}} \right)^2} + {\left( {{}^{20}{C_1}} \right)^2} + {\left( {{}^{20}{C_2}} \right)^2} + .... + {\left( {{}^{20}{C_{20}}} \right)^2}$$

= 40C20

Using the formula :

$${\left( {{}^n{C_0}} \right)^2} + {\left( {{}^n{C_1}} \right)^2} + {\left( {{}^n{C_2}} \right)^2} + .... + {\left( {{}^n{C_n}} \right)^2} = {}^{2n}{C_n}$$
2

### JEE Main 2021 (Online) 26th August Morning Shift

If $${{}^{20}{C_r}}$$ is the co-efficient of xr in the expansion of (1 + x)20, then the value of $$\sum\limits_{r = 0}^{20} {{r^2}.{}^{20}{C_r}}$$ is equal to :
A
$$420 \times {2^{19}}$$
B
$$380 \times {2^{19}}$$
C
$$380 \times {2^{18}}$$
D
$$420 \times {2^{18}}$$

## Explanation

$$\sum\limits_{r = 0}^{20} {{r^2}.{}^{20}{C_r}}$$

$$\sum {(4(r - 1) + r).{}^{20}{C_r}}$$

$$\sum {r(r - 1).{{20 \times 19} \over {r(r - 1)}}} .{}^{18}{C_r} + r.{{20} \over r}.\sum {{}^{19}{C_{r - 1}}}$$

$$\Rightarrow 20 \times {19.2^{18}} + {20.2^{19}}$$

$$\Rightarrow 420 \times {2^{18}}$$
3

### JEE Main 2021 (Online) 27th July Morning Shift

If the coefficients of x7 in $${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$$ and x$$-$$7 in $${\left( {{x} - {1 \over {bx^2}}} \right)^{11}}$$, b $$\ne$$ 0, are equal, then the value of b is equal to :
A
2
B
$$-$$1
C
1
D
$$-$$2

## Explanation

Coefficient of x7 in $${\left( {{x^2} + {1 \over {bx}}} \right)^{11}}$$ :

General Term = $${}^{11}{C_r}{({x^2})^{11 - r}}.{\left( {{1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{x^{22 - 3r}}.{1 \over {{b^r}}}$$

$$22 - 3r = 7$$

$$r = 5$$

$$\therefore$$ Required Term = $${}^{11}{C_5}.{1 \over {{b^5}}}.{x^7}$$

Coefficient of x$$-$$7 in $${\left( {x - {1 \over {b{x^2}}}} \right)^{11}}$$ :

General Term = $${}^{11}{C_r}{(x)^{11 - r}}.{\left( { - {1 \over {b{x^2}}}} \right)^r}$$

= $${}^{11}{C_r}{x^{11 - 3r}}.{{{{( - 1)}^r}} \over {{b^r}}}$$

$$11 - 3r = - 7$$ $$\therefore$$ $$r = 6$$

$$\therefore$$ Required Term = $${}^{11}{C_6}.{1 \over {{b^6}}}{x^{ - 7}}$$

According to the question,

$${}^{11}{C_5}.{1 \over {{b^5}}} = {}^{11}{C_6}.{1 \over {{b^6}}}$$

Since, b $$\ne$$ 0 $$\therefore$$ b = 1
4

### JEE Main 2021 (Online) 25th July Evening Shift

The lowest integer which is greater

than $${\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$$ is ______________.
A
3
B
4
C
2
D
1

## Explanation

Let $$P = {\left( {1 + {1 \over {{{10}^{100}}}}} \right)^{{{10}^{100}}}}$$

Let $$x = {10^{100}}$$

$$\Rightarrow P = {\left( {1 + {1 \over x}} \right)^x}$$

$$\Rightarrow P = 1 + (x)\left( {{1 \over x}} \right) + {{(x)(x - 1)} \over {\left| \!{\underline {\, 2 \,}} \right. }}.{1 \over {{x^2}}} + {{(x)(x - 1)(x - 2)} \over {\left| \!{\underline {\, 3 \,}} \right. }}.{1 \over {{x^3}}} + ....$$

(upto 10100 + 1 terms)

$$\Rightarrow P = 1 + 1 + \left( {{1 \over {\left| \!{\underline {\, 2 \,}} \right. }} - {1 \over {\left| \!{\underline {\, 2 \,}} \right. {x^2}}}} \right) + \left( {{1 \over {\left| \!{\underline {\, 3 \,}} \right. }} - ...} \right) + ...$$ so on

$$\Rightarrow P = 2 + \left( {Positive\,value\,less\,than\,{1 \over {\left| \!{\underline {\, 2 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 3 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 4 \,}} \right. }} + ...} \right)$$

Also $$e = 1 + {1 \over {\left| \!{\underline {\, 1 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 2 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 3 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 4 \,}} \right. }} + ...$$

$$\Rightarrow {1 \over {\left| \!{\underline {\, 2 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 3 \,}} \right. }} + {1 \over {\left| \!{\underline {\, 4 \,}} \right. }} + ... = e - 2$$

$$\Rightarrow$$ P = 2 + (positive value less than e $$-$$ 2)

$$\Rightarrow$$ P $$\in$$ (2, 3)

$$\Rightarrow$$ least integer value of P is 3

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