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Animated Solution for Mathematics - Binomial Theorem: Let $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}$ , then $a_{1} + a_{3} + a_{5} + \dots + a_{37}$ is equal to

Q: Wait, why do we subtract $f(-1)$ from $f(1)$? Why not add them?

Great question! When you expand $f(x) = a_0 + a_1x + a_2x^2 + \dots$, $f(1)$ gives you $a_0 + a_1 + a_2 + a_3 + \dots$, and $f(-1)$ gives you $a_0 - a_1 + a_2 - a_3 + \dots$. If you add them, the odd terms ($a_1, a_3, \dots$) cancel out, leaving you with the even terms. By subtracting them, the even terms cancel out, and the odd terms add up to $2(a_1 + a_3 + \dots)$. It is a classic algebraic trick to isolate parity!

Q: How did I know the degree of the polynomial is 40?

Look at the highest power term inside the bracket: $2x^2$. When you raise this to the power of 20, you get $(2x^2)^{20} = 2^{20} \cdot x^{40}$. Since this is the highest power of $x$ in the expansion, the degree of the polynomial is indeed 40. Always check the highest power before assuming the range of your coefficients!

Select Answer:

Progressive Hints

Hint 1

Hint 1: Look at the structure of the polynomial. When you see a sum of coefficients with alternating signs or specific parity (odd/even), your first instinct should be to evaluate the function at

x = 1

and

x = - 1

. How do these two values interact to isolate the odd-indexed terms?

Hint 2

Hint 2: Be careful! The standard formula for the sum of odd coefficients,

\frac{f ( 1 ) - f ( - 1 )}{2}

, gives you the sum of all odd coefficients up to the highest degree of the polynomial. Check the degree of

(1 + x + 2 x^{2})^{20}

. Does the question ask for the sum up to the very last term, or is there a missing piece?

Hint 3

Hint 3: You have successfully isolated the sum of all odd coefficients up to

a_{39}

. Now, to get the sum up to

a_{37}

, you must subtract the value of

a_{39}

. Use the multinomial expansion formula:

\frac{20 !}{p ! q ! r !} (1)^{p} (x)^{q} (2 x^{2})^{r}

. Set the exponent of

x

to 39 and solve for the non-negative integers

p, q, r

to find the coefficient

a_{39}

Visualized Solution

Understanding the Expansion

Given: $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}$
Target: $S = a_{1} + a_{3} + a_{5} + \dots + a_{37}$

Strategy for Odd Coefficients

Let $f (x) = (1 + x + 2 x^{2})^{20}$
Standard sum of all odd coefficients: $\sum a_{odd} = \frac{f ( 1 ) - f ( - 1 )}{2}$

Substitute $x = 1$

Put $x = 1$ in $f (x)$ :
$f (1) = (1 + 1 + 2 (1)^{2})^{20}$
$f (1) = 4^{20} = 2^{40}$
$2^{40} = a_{0} + a_{1} + a_{2} + \dots + a_{40}$

Substitute $x = - 1$

Put $x = - 1$ in $f (x)$ :
$f (- 1) = (1 - 1 + 2 (- 1)^{2})^{20}$
$f (- 1) = 2^{20}$
$2^{20} = a_{0} - a_{1} + a_{2} - a_{3} + \dots + a_{40}$

Isolating Odd Coefficients

Subtract $f (- 1)$ from $f (1)$ :
$f (1) - f (- 1) = 2^{40} - 2^{20}$
$2 (a_{1} + a_{3} + \dots + a_{39}) = 2^{40} - 2^{20}$
$a_{1} + a_{3} + \dots + a_{39} = 2^{39} - 2^{19}$

Identifying the Target Sum

Total odd sum: $a_{1} + a_{3} + \dots + a_{37} + a_{39} = 2^{19} (2^{20} - 1)$
Target sum: $S = a_{1} + a_{3} + \dots + a_{37}$
Therefore, $S = 2^{19} (2^{20} - 1) - a_{39}$

Finding $a_{39}$ via Multinomial Theorem

General term of $(1 + x + 2 x^{2})^{20}$ :
$\frac{20 !}{p ! q ! r !} (1)^{p} (x)^{q} (2 x^{2})^{r} = \frac{20 !}{p ! q ! r !} 2^{r} x^{q + 2 r}$
Constraints: $p + q + r = 20$ and $q + 2 r = 39$

Solving for $p, q, r$

We need $q + 2 r = 39$ and $p + q + r = 20$
If $r = 20 ⟹ q + 40 = 39 ⟹ q = - 1$ (Invalid)
If $r = 19 ⟹ q + 38 = 39 ⟹ q = 1$
Then $p = 20 - 19 - 1 = 0$
Valid solution: $(p, q, r) = (0, 1, 19)$

Calculating $a_{39}$

Substitute $(p, q, r) = (0, 1, 19)$ into the general term:
$a_{39} = \frac{20 !}{0 ! 1 ! 19 !} \cdot 2^{19}$
$a_{39} = 20 \cdot 2^{19}$

Final Calculation

$S = 2^{19} (2^{20} - 1) - a_{39}$
$S = 2^{19} (2^{20} - 1) - 20 \cdot 2^{19}$
Factor out $2^{19}$ :
$S = 2^{19} (2^{20} - 1 - 20)$
$S = 2^{19} (2^{20} - 21)$

00:00 / 00:00

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

When dealing with a trinomial expansion like

f (x) = (1 + x + 2 x^{2})^{20}

, brute force is rarely the intended path. We define the expansion as:

f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}

Our objective is to determine the sum

S = a_{1} + a_{3} + a_{5} + \dots + a_{37}

. This requires isolating the odd-indexed coefficients while excluding the term

a_{39}

The Parity Trick

We utilize the properties of functional evaluation to extract the sum of odd-indexed coefficients. First, we evaluate the function at

x = 1

f (1) = (1 + 1 + 2 (1)^{2})^{20} = 4^{20} = 2^{40} = a_{0} + a_{1} + a_{2} + \dots + a_{40}

Next, we evaluate at

x = - 1

to introduce alternating signs:

f (- 1) = (1 - 1 + 2 (- 1)^{2})^{20} = 2^{20} = a_{0} - a_{1} + a_{2} - a_{3} + \dots + a_{40}

By subtracting these two equations, the even-indexed terms cancel out, leaving us with twice the sum of the odd-indexed terms:

f (1) - f (- 1) = 2 (a_{1} + a_{3} + \dots + a_{39}) = 2^{40} - 2^{20}

Dividing by 2, we obtain the sum of all odd coefficients up to

a_{39}

a_{1} + a_{3} + \dots + a_{39} = 2^{39} - 2^{19} = 2^{19} (2^{20} - 1)

The Multinomial Rescue

The sum calculated above includes

a_{39}

, which is not part of our target sum

S

. We must isolate

a_{39}

using the Multinomial Theorem. The general term is:

T = \frac{20 !}{p ! q ! r !} (1)^{p} (x)^{q} (2 x^{2})^{r} = \frac{20 !}{p ! q ! r !} 2^{r} x^{q + 2 r}

We are constrained by

p + q + r = 20

and

q + 2 r = 39

. Testing values for

r

, we find that

r = 19

yields

q = 1

and

p = 0

. This is the only valid integer solution for the power 39.

Calculating the coefficient

a_{39}

a_{39} = \frac{20 !}{0 ! 1 ! 19 !} \cdot 2^{19} = 20 \cdot 2^{19}

Final Calculation

To find the target sum

S

, we subtract

a_{39}

from the previously derived total:

S = (a_{1} + a_{3} + \dots + a_{39}) - a_{39}

S = 2^{19} (2^{20} - 1) - 20 \cdot 2^{19}

Factoring out

2^{19}

, we arrive at the final result:

S = 2^{19} (2^{20} - 1 - 20)

$S = 2^{19} (2^{20} - 21)$

Common Traps & FAQs

Wait, why do we subtract

f (- 1)

from

f (1)

? Why not add them?

Great question! When you expand

f (x) = a_{0} + a_{1} x + a_{2} x^{2} + \dots

f (1)

gives you

a_{0} + a_{1} + a_{2} + a_{3} + \dots

, and

f (- 1)

gives you

a_{0} - a_{1} + a_{2} - a_{3} + \dots

. If you add them, the odd terms (

a_{1}, a_{3}, \dots

) cancel out, leaving you with the even terms. By subtracting them, the even terms cancel out, and the odd terms add up to

2 (a_{1} + a_{3} + \dots)

. It is a classic algebraic trick to isolate parity!

How did I know the degree of the polynomial is 40?

Look at the highest power term inside the bracket:

2 x^{2}

. When you raise this to the power of 20, you get

(2 x^{2})^{20} = 2^{20} \cdot x^{40}

. Since this is the highest power of

x

in the expansion, the degree of the polynomial is indeed 40. Always check the highest power before assuming the range of your coefficients!

Why can't I just use the Binomial Theorem here?

The Binomial Theorem applies to expressions with two terms, like

(a + b)^{n}

. Here, we have a trinomial:

(1 + x + 2 x^{2})^{20}

. While you could group terms like

(1 + (x + 2 x^{2}))^{20}

, it gets messy quickly. The Multinomial Theorem is the generalized tool for this, allowing us to handle any number of terms inside the bracket by using the general term formula

\frac{n !}{p ! q ! r !} a^{p} b^{q} c^{r}

Animated Solution for Mathematics - Binomial Theorem: Let $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}$ , then $a_{1} + a_{3} + a_{5} + \dots + a_{37}$ is equal to

Select Answer:

Visualized Solution

Understanding the Expansion

Strategy for Odd Coefficients

Substitute $x = 1$

Substitute $x = - 1$

Isolating Odd Coefficients

Identifying the Target Sum

Finding $a_{39}$ via Multinomial Theorem

Solving for $p, q, r$

Calculating $a_{39}$

Final Calculation

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Parity Trick

The Multinomial Rescue

Final Calculation

Similar Questions

Let $(1 + x + x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}$ . If $(a_{1} + a_{3} + a_{5} + \dots + a_{19}) - 11 a_{2} = 121 k$ , then $k$ is equal to .

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

If $^{20} C_{r}$ is the co-efficient of $x^{r}$ in the expansion of $(1 + x)^{20}$ , then the value of $\sum_{r = 0}^{20} r^{2}^{20} C_{r}$ is equal to :

For $r = 0, 1, ..., 10$ , let $A_{r}, B_{r}$ and $C_{r}$ denote, respectively, the coefficient of $x^{r}$ in the expansions of $(1 + x)^{10}, (1 + x)^{20}$ and $(1 + x)^{30}$ . Then $\sum_{r = 1}^{10} A_{r} (B_{10} B_{r} - C_{10} A_{r})$ is equal to

The sum of the series $2 .^{20} C_{0} + 5 .^{20} C_{1} + 8 .^{20} C_{2} + 11 .^{20} C_{3} + ... + 62 .^{20} C_{20}$ is equal to :

If $1^{2} \cdot (^{15} C_{1}) + 2^{2} \cdot (^{15} C_{2}) + 3^{2} \cdot (^{15} C_{3}) + \dots + 1 5^{2} \cdot (^{15} C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where $m, n, k \in N$ , then $m + n + k$ is equal to :

$\sum_{k = 0}^{20} (^{20} C_{k})^{2}$ is equal to :

The coefficient of $x^{48}$ in $(1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 100 (1 + x)^{100}$ is equal to

Let $n$ be a positive integer and $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{2 n} x^{2 n}$ . Show that $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

If $^{20} C_{1} + (2^{2})^{20} C_{2} + (3^{2})^{20} C_{3} + \dots + (2 0^{2})^{20} C_{20} = A (2^{β})$ , then the ordered pair $(A, β)$ is equal to:

Animated Solution for Mathematics - Binomial Theorem: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (1+x+2x2)20=a0​+a1​x+a2​x2+⋯+a40​x40, then a1​+a3​+a5​+⋯+a37​ is equal to

Select Answer:

Visualized Solution

Understanding the Expansion

Strategy for Odd Coefficients

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=1

Substitute .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } x=−1

Isolating Odd Coefficients

Identifying the Target Sum

Finding .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a39​ via Multinomial Theorem

Solving for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } p,q,r

Calculating .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a39​

Final Calculation

The Sigma Insight: Properties of Binomial Coefficients

Analyzing the Setup

The Parity Trick

The Multinomial Rescue

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let (1+x+x2)10=a0​+a1​x+a2​x2+⋯+a20​x20. If (a1​+a3​+a5​+⋯+a19​)−11a2​=121k, then k is equal to .

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The value of (21C1​−10C1​)+(21C2​−10C2​)+(21C3​−10C3​)+(21C4​−10C4​)+⋯+(21C10​−10C10​) is:

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 20Cr​ is the co-efficient of xr in the expansion of (1+x)20, then the value of ∑r=020​r220Cr​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the series 2.20C0​+5.20C1​+8.20C2​+11.20C3​+...+62.20C20​ is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 12⋅(15C1​)+22⋅(15C2​)+32⋅(15C3​)+⋯+152⋅(15C15​)=2m⋅3n⋅5k, where m,n,k∈N, then m+n+k is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } ∑k=020​(20Ck​)2 is equal to :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The coefficient of x48 in (1+x)+2(1+x)2+3(1+x)3+⋯+100(1+x)100 is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let n be a positive integer and (1+x+x2)n=a0​+a1​x+a2​x2+....+a2n​x2n. Show that a02​−a12​+a22​−....+a2n2​=an​.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } If 20C1​+(22)20C2​+(32)20C3​+⋯+(202)20C20​=A(2β), then the ordered pair (A,β) is equal to:

Animated Solution for Mathematics - Binomial Theorem: Let $(1 + x + 2 x^{2})^{20} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{40} x^{40}$ , then $a_{1} + a_{3} + a_{5} + \dots + a_{37}$ is equal to

Substitute $x = 1$

Substitute $x = - 1$

Finding $a_{39}$ via Multinomial Theorem

Solving for $p, q, r$

Calculating $a_{39}$

Let $(1 + x + x^{2})^{10} = a_{0} + a_{1} x + a_{2} x^{2} + \dots + a_{20} x^{20}$ . If $(a_{1} + a_{3} + a_{5} + \dots + a_{19}) - 11 a_{2} = 121 k$ , then $k$ is equal to .

The value of $(^{21} C_{1} -^{10} C_{1}) + (^{21} C_{2} -^{10} C_{2}) + (^{21} C_{3} -^{10} C_{3}) + (^{21} C_{4} -^{10} C_{4}) + \dots + (^{21} C_{10} -^{10} C_{10})$ is:

If $^{20} C_{r}$ is the co-efficient of $x^{r}$ in the expansion of $(1 + x)^{20}$ , then the value of $\sum_{r = 0}^{20} r^{2}^{20} C_{r}$ is equal to :

The sum of the series $2 .^{20} C_{0} + 5 .^{20} C_{1} + 8 .^{20} C_{2} + 11 .^{20} C_{3} + ... + 62 .^{20} C_{20}$ is equal to :

If $1^{2} \cdot (^{15} C_{1}) + 2^{2} \cdot (^{15} C_{2}) + 3^{2} \cdot (^{15} C_{3}) + \dots + 1 5^{2} \cdot (^{15} C_{15}) = 2^{m} \cdot 3^{n} \cdot 5^{k}$ , where $m, n, k \in N$ , then $m + n + k$ is equal to :

$\sum_{k = 0}^{20} (^{20} C_{k})^{2}$ is equal to :

The coefficient of $x^{48}$ in $(1 + x) + 2 (1 + x)^{2} + 3 (1 + x)^{3} + \dots + 100 (1 + x)^{100}$ is equal to

Let $n$ be a positive integer and $(1 + x + x^{2})^{n} = a_{0} + a_{1} x + a_{2} x^{2} + .... + a_{2 n} x^{2 n}$ . Show that $a_{0}^{2} - a_{1}^{2} + a_{2}^{2} - .... + a_{2 n}^{2} = a_{n}$ .

If $^{20} C_{1} + (2^{2})^{20} C_{2} + (3^{2})^{20} C_{3} + \dots + (2 0^{2})^{20} C_{20} = A (2^{β})$ , then the ordered pair $(A, β)$ is equal to: