JEE Advanced 2006

LEVELJEE Advanced

Animated Solution for Mathematics - Sequence and Series: If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n}$ for all $n \geq n_{0}$ .

Q: Wait, why can't we just use the sum to infinity formula for $a_n$?

That is a classic trap! The sum to infinity formula $S_{\infty} = \frac{a}{1-r}$ is only valid when the number of terms is infinite. Here, $n$ is a finite natural number. We are dealing with a partial sum, so we must use the finite sum formula $S_n = \frac{a(1-r^n)}{1-r}$.

Q: Why does the inequality sign flip when I multiply by -1 in the last step?

This is a fundamental rule of inequalities. Whenever you multiply or divide an inequality by a negative number, the direction of the inequality must reverse. Think of it like a mirror reflection on the number line; if $x > y$, then $-x < -y$.

Q: Why do we have to check even and odd cases for $n$?

Because the common ratio is negative ($r = -\frac{3}{4}$), the term $(-\frac{3}{4})^n$ oscillates. If $n$ is even, the result is positive. If $n$ is odd, the result is negative. Since the inequality behaves differently for positive and negative values, we must split the analysis to ensure we find the correct $n_0$.

Enter Numerical Value:

Progressive Hints

Hint 1

Hint 1: The Spark. Visualize the series

a_{n}

as a sum of a Geometric Progression. The alternating signs are the key. Don't let the

(- 1)^{n - 1}

term intimidate you; it simply means the common ratio

r

is negative. Focus on the structure

a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} - \dots

Hint 2

Hint 2: The Arsenal. You are given

b_{n} = 1 - a_{n}

. Instead of calculating

b_{n}

and

a_{n}

separately, substitute this definition directly into the inequality

b_{n} > a_{n}

. This transforms the problem into

1 - a_{n} > a_{n}

, which simplifies to

a_{n} < \frac{1}{2}

. This is a much cleaner path.

Hint 3

Hint 3: The Execution. Once you have

a_{n} < \frac{1}{2}

, substitute the formula for the sum of a G.P. into this inequality. You will arrive at an expression involving

(- \frac{3}{4})^{n}

. Remember that the behavior of this term changes drastically depending on whether

n

is even or odd. Analyze these two cases separately to find the threshold

n_{0}

Visualized Solution

Identify the Series $a_{n}$

Given series: $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$
This is a Geometric Progression (G.P.) with:
First term $a = \frac{3}{4}$
Common ratio $r = - \frac{3}{4}$

Apply G.P. Sum Formula

Sum of $n$ terms of a G.P.: $S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}$
Substituting $a = \frac{3}{4}$ and $r = - \frac{3}{4}$ :
$a_{n} = \frac{\frac{3}{4} ( 1 - ( - \frac{3}{4} ) ^{n} )}{1 - ( - \frac{3}{4} )}$

Simplify the Expression for $a_{n}$

Denominator: $1 - (- \frac{3}{4}) = 1 + \frac{3}{4} = \frac{7}{4}$
$a_{n} = \frac{\frac{3}{4}}{\frac{7}{4}} (1 - (- \frac{3}{4})^{n})$
$a_{n} = \frac{3}{7} (1 - (- \frac{3}{4})^{n})$

Define $b_{n}$ in Terms of $n$

Given: $b_{n} = 1 - a_{n}$
$b_{n} = 1 - [\frac{3}{7} (1 - (- \frac{3}{4})^{n})]$
$b_{n} = 1 - \frac{3}{7} + \frac{3}{7} (- \frac{3}{4})^{n}$
$b_{n} = \frac{4}{7} + \frac{3}{7} (- \frac{3}{4})^{n}$

Set up the Inequality $b_{n} > a_{n}$

Condition: $b_{n} > a_{n}$
Substitute $b_{n} = 1 - a_{n}$ :
$1 - a_{n} > a_{n}$
$1 > 2 a_{n} \Rightarrow a_{n} < \frac{1}{2}$

Simplify the Inequality

Substitute $a_{n}$ into the inequality:
$\frac{3}{7} (1 - (- \frac{3}{4})^{n}) < \frac{1}{2}$
$1 - (- \frac{3}{4})^{n} < \frac{7}{6}$
$- (- \frac{3}{4})^{n} < \frac{7}{6} - 1 = \frac{1}{6}$
$(- \frac{3}{4})^{n} > - \frac{1}{6}$

Analyze Even $n$ Case

Case 1: $n$ is even
$(- \frac{3}{4})^{n} = (\frac{3}{4})^{n}$ , which is always positive.
Since a positive number is always $> - \frac{1}{6}$ , the condition $b_{n} > a_{n}$ is always true for all even $n$ .

Analyze Odd $n$ Case

Case 2: $n$ is odd
$(- \frac{3}{4})^{n} = - (\frac{3}{4})^{n}$
Inequality becomes: $- (\frac{3}{4})^{n} > - \frac{1}{6}$
Multiplying by $- 1$ flips the sign: $(\frac{3}{4})^{n} < \frac{1}{6}$

Test Values for Odd $n$

Testing odd values of $n$ :
For $n = 1 : \frac{3}{4} = 0.75 \neq < 0.166$ (False)
For $n = 3 : \frac{27}{64} \approx 0.421 \neq < 0.166$ (False)
For $n = 5 : \frac{243}{1024} \approx 0.237 \neq < 0.166$ (False)
For $n = 7 : \frac{2187}{16384} \approx 0.133 < 0.166$ (True)

Conclusion: Find $n_{0}$

We found:
$n = 5$ is False ( $b_{5} < a_{5}$ )
$n = 6$ is True ( $n$ is even)
$n = 7$ is True ( $(\frac{3}{4})^{7} < \frac{1}{6}$ )
Since $(\frac{3}{4})^{n}$ is a decreasing function, it holds for all $n \geq 7$ (odd) and $n \geq 6$ (even).
The least natural number $n_{0}$ is 6.

00:00 / 00:00

The Sigma Insight: Geometric Progression (G.P.)

Solution Diagram

Analyzing the Setup

The sequence is defined as

a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}

This is a Geometric Progression (G.P.) where the first term is

a = \frac{3}{4}

and the common ratio is

r = - \frac{3}{4}

The Algebraic Shortcut

We are tasked with finding

n_{0}

such that

b_{n} > a_{n}

, where

b_{n} = 1 - a_{n}

Substituting the definition of

b_{n}

into the inequality, we get

1 - a_{n} > a_{n}

This simplifies to

1 > 2 a_{n}

, or equivalently:

a_{n} < \frac{1}{2}

The Summation

Using the finite sum formula for a G.P.,

S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}

, we substitute our values:

a_{n} = \frac{\frac{3}{4} ( 1 - ( - \frac{3}{4} ) ^{n} )}{1 - ( - \frac{3}{4} )}

The denominator simplifies to

1 + \frac{3}{4} = \frac{7}{4}

. The fours cancel out, yielding:

a_{n} = \frac{3}{7} (1 - (- \frac{3}{4})^{n})

Now, we substitute this into our inequality

a_{n} < \frac{1}{2}

\frac{3}{7} (1 - (- \frac{3}{4})^{n}) < \frac{1}{2}

Multiplying by

\frac{7}{3}

and rearranging, we isolate the term involving

n

1 - (- \frac{3}{4})^{n} < \frac{7}{6}

- (- \frac{3}{4})^{n} < \frac{1}{6}

(- \frac{3}{4})^{n} > - \frac{1}{6}

The Bifurcation

Even vs. Odd

n

is even,

(- \frac{3}{4})^{n}

is positive. Since any positive number is greater than

- \frac{1}{6}

, the condition holds for all even

n

n

is odd,

(- \frac{3}{4})^{n} = - (\frac{3}{4})^{n}

. The inequality becomes:

- (\frac{3}{4})^{n} > - \frac{1}{6} \Rightarrow (\frac{3}{4})^{n} < \frac{1}{6}

We test odd values for

n

: - For

n = 1

, $0.75 ot< 0.166$ - For

n = 3

, $0.42 ot< 0.166$ - For

n = 5

, $0.23 ot< 0.166$ - For

n = 7

(\frac{3}{4})^{7} \approx 0.133 < 0.166

Final Conclusion

The condition holds for all even

n

and for all odd

n \geq 7

Checking the sequence of integers:

n = 1

(False),

n = 2

(True),

n = 3

(False),

n = 4

(True),

n = 5

(False),

n = 6

(True),

n = 7

(True).

The least natural number

n_{0}

such that the condition holds for all

n \geq n_{0}

is $n_{0} = 6$ .

Common Traps & FAQs

Wait, why can't we just use the sum to infinity formula for

a_{n}

That is a classic trap! The sum to infinity formula

S_{\infty} = \frac{a}{1 - r}

is only valid when the number of terms is infinite. Here,

n

is a finite natural number. We are dealing with a partial sum, so we must use the finite sum formula

S_{n} = \frac{a ( 1 - r ^{n} )}{1 - r}

Why does the inequality sign flip when I multiply by -1 in the last step?

This is a fundamental rule of inequalities. Whenever you multiply or divide an inequality by a negative number, the direction of the inequality must reverse. Think of it like a mirror reflection on the number line; if

x > y

, then

- x < - y

Why do we have to check even and odd cases for

n

Because the common ratio is negative (

r = - \frac{3}{4}

), the term

(- \frac{3}{4})^{n}

oscillates. If

n

is even, the result is positive. If

n

is odd, the result is negative. Since the inequality behaves differently for positive and negative values, we must split the analysis to ensure we find the correct

n_{0}

Animated Solution for Mathematics - Sequence and Series: If $a_{n} = \frac{3}{4} - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} + \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $b_{n} = 1 - a_{n}$ , then find the least natural number $n_{0}$ such that $b_{n} > a_{n}$ for all $n \geq n_{0}$ .

Enter Numerical Value:

Visualized Solution

Identify the Series $a_{n}$

Apply G.P. Sum Formula

Simplify the Expression for $a_{n}$

Define $b_{n}$ in Terms of $n$

Set up the Inequality $b_{n} > a_{n}$

Simplify the Inequality

Analyze Even $n$ Case

Analyze Odd $n$ Case

Test Values for Odd $n$

Conclusion: Find $n_{0}$

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Algebraic Shortcut

The Summation

The Bifurcation

Final Conclusion

Similar Questions

Let, $A_{n} = (\frac{3}{4}) - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} - \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $B_{n} = 1 - A_{n}$ . Then, the least odd natural number p, so that $B_{n} > A_{n}$ for all $n \geq p$ , is :-

Let $A_{1}, A_{2}, A_{3}, \dots$ be squares such that for each $n \geq 1$ , the length of the side of $A_{n}$ equals the length of diagonal of $A_{n + 1}$ . If the length of $A_{1}$ is 12 cm, then the smallest value of $n$ for which area of $A_{n}$ is less than one, is

Find the natural number 'a' for which $\sum_{k = 1}^{n} f (a + k) = 16 (2^{n} - 1)$ , where the function 'f' satisfies the relation $f (x + y) = f (x) f (y)$ for all natural numbers $x, y$ and further $f (1) = 2$ .

Let $A = {x \in R : [x + 3] + [x + 4] \leq 3}, B = {x \in R : 3^{x} (\sum_{r = 1}^{\infty} \frac{3}{1 0 ^{r}})^{x - 3} < 3^{- 3 x}}$ , where $[t]$ denotes greatest integer function. Then,

Let $A_{1}, A_{2}, A_{3}, \dots$ be an increasing geometric progression of positive real numbers. If $A_{1} A_{3} A_{5} A_{7} = \frac{1}{1296}$ and $A_{2} + A_{4} = \frac{7}{36}$ , then, the value of $A_{6} + A_{8} + A_{10}$ is equal to

If for $x, y \in R, x > 0$ \\ $y = lo g_{10} x + lo g_{10} x^{1/3} + lo g_{10} x^{1/9} + \dots$ upto $\infty$ terms\\ and $\frac{2 + 4 + 6 + \dots + 2 y}{3 + 6 + 9 + \dots + 3 y} = \frac{4}{l o g _{10} x}$ , then the ordered pair $(x, y)$ is equal to :

Let $S_{1}, S_{2}, \dots$ be squares such that for each $n \geq 1$ , the length of a side of $S_{n}$ equals the length of a diagonal of $S_{n + 1}$ . If the length of a side of $S_{1}$ is 10 cm, then for which of the following values of $n$ is the area of $S_{n}$ less than 1 sq. cm?

Let $\sum_{k = 1}^{10} f (a + k) = 16 (2^{10} - 1)$ , where the function $f$ satisfies $f (x + y) = f (x) f (y)$ for all natural numbers $x, y$ and $f (1) = 2$ . then the natural number 'a' is

Let $a_{1} = 1$ and for $n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Then $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$ is equal to ......... .

Enter Numerical Value:

Visualized Solution

Identify the Series .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​

Apply G.P. Sum Formula

Simplify the Expression for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } an​

Define .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn​ in Terms of .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Set up the Inequality .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } bn​>an​

Simplify the Inequality

Analyze Even .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n Case

Analyze Odd .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n Case

Test Values for Odd .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n

Conclusion: Find .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n0​

The Sigma Insight: Geometric Progression (G.P.)

Analyzing the Setup

The Algebraic Shortcut

The Summation

The Bifurcation

Final Conclusion

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let, An​=(43​)−(43​)2+(43​)3−⋯+(−1)n−1(43​)n and Bn​=1−An​. Then, the least odd natural number p, so that Bn​>An​ for all n≥p, is :-

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let A={x∈R:[x+3]+[x+4]≤3},B={x∈R:3x(∑r=1∞​10r3​)x−3<3−3x}, where [t] denotes greatest integer function. Then,

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If for x,y∈R,x>0\\ y=log10​x+log10​x1/3+log10​x1/9+… upto ∞ terms\\ and 3+6+9+⋯+3y2+4+6+⋯+2y​=log10​x4​, then the ordered pair (x,y) is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let ∑k=110​f(a+k)=16(210−1), where the function f satisfies f(x+y)=f(x)f(y) for all natural numbers x,y and f(1)=2. then the natural number 'a' is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let a1​=1 and for n≥1,an+1​=21​an​+n2(n+1)2n2−2n−1​. Then ∑n=1∞​(an​−n22​) is equal to ......... .

Identify the Series $a_{n}$

Simplify the Expression for $a_{n}$

Define $b_{n}$ in Terms of $n$

Set up the Inequality $b_{n} > a_{n}$

Analyze Even $n$ Case

Analyze Odd $n$ Case

Test Values for Odd $n$

Conclusion: Find $n_{0}$

Let, $A_{n} = (\frac{3}{4}) - (\frac{3}{4})^{2} + (\frac{3}{4})^{3} - \dots + (- 1)^{n - 1} (\frac{3}{4})^{n}$ and $B_{n} = 1 - A_{n}$ . Then, the least odd natural number p, so that $B_{n} > A_{n}$ for all $n \geq p$ , is :-

Let $A = {x \in R : [x + 3] + [x + 4] \leq 3}, B = {x \in R : 3^{x} (\sum_{r = 1}^{\infty} \frac{3}{1 0 ^{r}})^{x - 3} < 3^{- 3 x}}$ , where $[t]$ denotes greatest integer function. Then,

If for $x, y \in R, x > 0$ \\ $y = lo g_{10} x + lo g_{10} x^{1/3} + lo g_{10} x^{1/9} + \dots$ upto $\infty$ terms\\ and $\frac{2 + 4 + 6 + \dots + 2 y}{3 + 6 + 9 + \dots + 3 y} = \frac{4}{l o g _{10} x}$ , then the ordered pair $(x, y)$ is equal to :

Let $\sum_{k = 1}^{10} f (a + k) = 16 (2^{10} - 1)$ , where the function $f$ satisfies $f (x + y) = f (x) f (y)$ for all natural numbers $x, y$ and $f (1) = 2$ . then the natural number 'a' is

Let $a_{1} = 1$ and for $n \geq 1, a_{n + 1} = \frac{1}{2} a_{n} + \frac{n ^{2} - 2 n - 1}{n ^{2} ( n + 1 ) ^{2}}$ . Then $\sum_{n = 1}^{\infty} (a_{n} - \frac{2}{n ^{2}})$ is equal to ......... .