JEE Main 2020 (7 January Shift 2)

LEVELJEE Main

Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^{7}$ in the expression $(1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + \dots + x^{10}$ is :

Q: Why does the $-x^{11}$ term disappear when we look for the coefficient of $x^7$?

Great question! After simplifying the series, we get $S = (1+x)^{11} - x^{11}$. We are looking for the coefficient of $x^7$. The term $-x^{11}$ is a polynomial of degree $11$. It only contains an $x^{11}$ term. It has absolutely no $x^7$ component. Therefore, its contribution to the coefficient of $x^7$ is zero. We only need to extract the $x^7$ term from the $(1+x)^{11}$ part.

Select Answer:

Visualized Solution

Observe the Pattern

Expression: $S = (1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + \dots + x^{10}$
Observe that the power of $(1 + x)$ decreases from $10$ to $0$ .
Observe that the power of $x$ increases from $0$ to $10$ .

Identify the Series Type

The expression is a Geometric Progression (G.P.).
Check ratio: $\frac{x ( 1 + x ) ^{9}}{( 1 + x ) ^{10}} = \frac{x}{1 + x}$
Check next ratio: $\frac{x ^{2} ( 1 + x ) ^{8}}{x ( 1 + x ) ^{9}} = \frac{x}{1 + x}$

Identify G.P. Parameters

First term $a = (1 + x)^{10}$
Common ratio $r = \frac{x}{1 + x}$
Number of terms $n = 11$ (since powers of $x$ range from $0$ to $10$ )

Apply G.P. Sum Formula

Sum of G.P. formula: $S = a \frac{1 - r ^{n}}{1 - r}$
Substitute values: $S = (1 + x)^{10} \frac{1 - ( \frac{x}{1 + x} ) ^{11}}{1 - \frac{x}{1 + x}}$

Simplify the Denominator

Simplify denominator: $1 - \frac{x}{1 + x} = \frac{( 1 + x ) - x}{1 + x}$
Denominator $= \frac{1}{1 + x}$

Combine Terms

Substitute back: $S = (1 + x)^{10} \cdot (1 + x) [1 - \frac{x ^{11}}{( 1 + x ) ^{11}}]$
$S = (1 + x)^{11} [1 - \frac{x ^{11}}{( 1 + x ) ^{11}}]$

Final Simplified Expression

Distribute: $S = (1 + x)^{11} \cdot 1 - (1 + x)^{11} \cdot \frac{x ^{11}}{( 1 + x ) ^{11}}$
Simplified Sum: $S = (1 + x)^{11} - x^{11}$

Identify the Target Coefficient

We need the coefficient of $x^{7}$ in $(1 + x)^{11} - x^{11}$ .
In $(1 + x)^{11}$ , general term $T_{r + 1} = (r 11) x^{r}$ .
For $x^{7}$ , set $r = 7$ . Coefficient is $(7 11)$ .
In $- x^{11}$ , the coefficient of $x^{7}$ is $0$ .

Calculate the Final Value

Coefficient $= (7 11) = (11 - 7 11) = (4 11)$
$(4 11) = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1}$
$(4 11) = 11 \times 10 \times 3 = 330$

Summary and Conclusion

Key Takeaway: Recognize G.P. patterns in binomial sums to simplify the expression before finding coefficients.
Final Answer: The coefficient of $x^{7}$ is 330.

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The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The given expression is

S = (1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + \dots + x^{10}

. At first glance, it appears to be a complex series, but we can identify it as a Geometric Progression (G.P.).

In this series, the first term is

a = (1 + x)^{10}

and the common ratio is

r = \frac{x}{1 + x}

. The total number of terms in this sequence is

n = 11

The Master Equation

We utilize the standard sum formula for a G.P., which is given by:

S = a \frac{1 - r ^{n}}{1 - r}

Substituting our identified values into the formula, we obtain:

S = (1 + x)^{10} [\frac{1 - ( \frac{x}{1 + x} ) ^{11}}{1 - \frac{x}{1 + x}}]

Simplifying the Expression

Focusing on the denominator, we simplify

1 - \frac{x}{1 + x}

1 - \frac{x}{1 + x} = \frac{( 1 + x ) - x}{1 + x} = \frac{1}{1 + x}

Since dividing by

\frac{1}{1 + x}

is equivalent to multiplying by

(1 + x)

, the expression for

S

becomes:

S = (1 + x)^{10} \cdot (1 + x) [1 - \frac{x ^{11}}{( 1 + x ) ^{11}}]

Distributing the

(1 + x)^{11}

term across the bracket, we arrive at the elegant result:

S = (1 + x)^{11} - x^{11}

Final Calculation

To find the coefficient of

x^{7}

in the expansion of

S

, we examine the binomial expansion of

(1 + x)^{11}

. The term containing

x^{7}

is given by:

(7 11) x^{7} = (4 11) x^{7}

Calculating the binomial coefficient:

(4 11) = \frac{11 \times 10 \times 9 \times 8}{4 \times 3 \times 2 \times 1} = 330

The coefficient of

x^{7}

in the expansion is 330.

Common Traps & FAQs

Wait, why is this a Geometric Progression? The ratio involves a variable

x

, not a constant number.

That is a brilliant observation! In a G.P., the common ratio

r

does not have to be a constant number; it can be a function of a variable. As long as the ratio between any two consecutive terms is the same, it satisfies the definition of a G.P. Here,

\frac{T _{2}}{T _{1}} = \frac{x ( 1 + x ) ^{9}}{( 1 + x ) ^{10}} = \frac{x}{1 + x}

, and

\frac{T _{3}}{T _{2}} = \frac{x ^{2} ( 1 + x ) ^{8}}{x ( 1 + x ) ^{9}} = \frac{x}{1 + x}

. Since the ratio is consistent, the G.P. rules apply perfectly.

How do I know there are exactly 11 terms? I get confused with the counting.

The easiest way to count terms in a sequence like this is to look at the exponent of one of the variables. The power of

x

starts at

0

(in the first term) and ends at

10

(in the last term). The number of integers from

0

10

inclusive is

10 - 0 + 1

, which equals

11

. Always use the 'last minus first plus one' rule to avoid off-by-one errors!

Why does the

- x^{11}

term disappear when we look for the coefficient of

x^{7}

Great question! After simplifying the series, we get

S = (1 + x)^{11} - x^{11}

. We are looking for the coefficient of

x^{7}

. The term

- x^{11}

is a polynomial of degree

11

. It only contains an

x^{11}

term. It has absolutely no

x^{7}

component. Therefore, its contribution to the coefficient of

x^{7}

is zero. We only need to extract the

x^{7}

term from the

(1 + x)^{11}

part.

Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^{7}$ in the expression $(1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + \dots + x^{10}$ is :

Select Answer:

Visualized Solution

Observe the Pattern

Identify the Series Type

Identify G.P. Parameters

Apply G.P. Sum Formula

Simplify the Denominator

Combine Terms

Final Simplified Expression

Identify the Target Coefficient

Calculate the Final Value

Summary and Conclusion

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Master Equation

Simplifying the Expression

Final Calculation

Similar Questions

The coefficient of $x^{7}$ in the expression $(1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + ... + x^{10}$ is :

The coefficient of $x^{7}$ in the expansion of $(1 - x - x^{2} + x^{3})^{6}$ is

The coefficient of $x^{9}$ in the expansion of $(1 + x) (1 + x^{2}) (1 + x^{3}) \dots (1 + x^{100})$ is \dots.

The coefficient of $x^{301}$ in $(1 + x)^{500} + x (1 + x)^{499} + x^{2} (1 + x)^{498} + \dots + x^{500}$ is:

The coefficient of $x^{2012}$ in the expansion of $(1 - x)^{2008} (1 + x + x^{2})^{2007}$ is equal to

The coefficient of $x^{2}$ in the expansion of the product $(2 - x^{2}) \cdot ((1 + 2 x + 3 x^{2})^{6} + (1 - 4 x^{2})^{6})$ is :

In the expansion of $(1 + x) (1 - x^{2}) (1 + \frac{3}{x} + \frac{3}{x ^{2}} + \frac{1}{x ^{3}})^{5}, x \neq = 0$ , the sum of the coefficient of $x^{3}$ and $x^{- 13}$ is equal to _______.

The coefficient of $x^{101}$ in the expression $(5 + x)^{500} + x (5 + x)^{499} + x^{2} (5 + x)^{498} + \dots + x^{500}$ , $x > 0$ , is

The coefficient of $x^{256}$ in the expansion of $(1 - x)^{101} (x^{2} + x + 1)^{100}$ is:

The coefficient of $x^{n}$ in expansion of $(1 + x) (1 - x)^{n}$ is

Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x7 in the expression (1+x)10+x(1+x)9+x2(1+x)8+⋯+x10 is :

Select Answer:

Visualized Solution

Observe the Pattern

Identify the Series Type

Identify G.P. Parameters

Apply G.P. Sum Formula

Simplify the Denominator

Combine Terms

Final Simplified Expression

Identify the Target Coefficient

Calculate the Final Value

Summary and Conclusion

The Sigma Insight: Binomial Expansion for Positive Integral Index

Analyzing the Setup

The Master Equation

Simplifying the Expression

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x7 in the expression (1+x)10+x(1+x)9+x2(1+x)8+...+x10 is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x7 in the expansion of (1−x−x2+x3)6 is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x9 in the expansion of (1+x)(1+x2)(1+x3)…(1+x100) is \dots.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x301 in (1+x)500+x(1+x)499+x2(1+x)498+⋯+x500 is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x2012 in the expansion of (1−x)2008(1+x+x2)2007 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x2 in the expansion of the product (2−x2)⋅((1+2x+3x2)6+(1−4x2)6) is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the expansion of (1+x)(1−x2)(1+x3​+x23​+x31​)5,x=0, the sum of the coefficient of x3 and x−13 is equal to _______.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x101 in the expression (5+x)500+x(5+x)499+x2(5+x)498+⋯+x500, x>0, is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x256 in the expansion of (1−x)101(x2+x+1)100 is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of xn in expansion of (1+x)(1−x)n is

Animated Solution for Mathematics - Binomial Theorem: The coefficient of $x^{7}$ in the expression $(1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + \dots + x^{10}$ is :

The coefficient of $x^{7}$ in the expression $(1 + x)^{10} + x (1 + x)^{9} + x^{2} (1 + x)^{8} + ... + x^{10}$ is :

The coefficient of $x^{7}$ in the expansion of $(1 - x - x^{2} + x^{3})^{6}$ is

The coefficient of $x^{9}$ in the expansion of $(1 + x) (1 + x^{2}) (1 + x^{3}) \dots (1 + x^{100})$ is \dots.

The coefficient of $x^{301}$ in $(1 + x)^{500} + x (1 + x)^{499} + x^{2} (1 + x)^{498} + \dots + x^{500}$ is:

The coefficient of $x^{2012}$ in the expansion of $(1 - x)^{2008} (1 + x + x^{2})^{2007}$ is equal to

The coefficient of $x^{2}$ in the expansion of the product $(2 - x^{2}) \cdot ((1 + 2 x + 3 x^{2})^{6} + (1 - 4 x^{2})^{6})$ is :

In the expansion of $(1 + x) (1 - x^{2}) (1 + \frac{3}{x} + \frac{3}{x ^{2}} + \frac{1}{x ^{3}})^{5}, x \neq = 0$ , the sum of the coefficient of $x^{3}$ and $x^{- 13}$ is equal to _______.

The coefficient of $x^{101}$ in the expression $(5 + x)^{500} + x (5 + x)^{499} + x^{2} (5 + x)^{498} + \dots + x^{500}$ , $x > 0$ , is

The coefficient of $x^{256}$ in the expansion of $(1 - x)^{101} (x^{2} + x + 1)^{100}$ is:

The coefficient of $x^{n}$ in expansion of $(1 + x) (1 - x)^{n}$ is