Why can't we just use the formula for even $n$ for odd $n$?

The even formula $S_n = \frac{n(n+1)^2}{2}$ is specifically derived for even $n$ because the even-positioned terms have that extra factor of 2. Applying it to odd $n$ ignores the structural difference of the series.

How do I know $T_n = n^2$ when $n$ is odd?

Look at the pattern: $T_1 = 1^2$, $T_3 = 3^2$, $T_5 = 5^2$. The term at position $n$ is simply the square of the position index $n$.

Is there a single formula for both?

While you could construct a piecewise function, in JEE, it is often more elegant and less error-prone to handle the cases separately as we did here.

JEE Advanced 1988

LEVELJEE Main

Animated Solution for Mathematics - Sequence and Series: The sum of the first $n$ terms of the series $1^{2} + 2. 2^{2} + 3^{2} + 2. 4^{2} + 5^{2} + 2. 6^{2} + \dots$ is $n (n + 1)^{2} /2$ , when $n$ is even. When $n$ is odd, the sum is \dots.

Visualized Solution

Analyze the Series Pattern

Given series: $1^{2} + 2 \cdot 2^{2} + 3^{2} + 2 \cdot 4^{2} + 5^{2} + 2 \cdot 6^{2} + \dots$
Let's observe the terms to find a hidden pattern.

Odd and Even Position Terms

Odd position terms: $T_{1} = 1^{2}$ , $T_{3} = 3^{2}$ , $T_{5} = 5^{2} \dots$
Even position terms: $T_{2} = 2 \cdot 2^{2}$ , $T_{4} = 2 \cdot 4^{2}$ , $T_{6} = 2 \cdot 6^{2} \dots$

Sum for Even Number of Terms

Given: For an even number of terms $n$ , the sum is $S_{n} = \frac{n ( n + 1 ) ^{2}}{2}$ .

Strategy for Odd $n$

We need to find the sum $S_{n}$ when $n$ is odd.
If $n$ is odd, then $(n - 1)$ must be even.

Splitting the Sum

We can write the sum of $n$ terms as:
$S_{n} = S_{n - 1} + T_{n}$
Here, $S_{n - 1}$ is the sum of the first $(n - 1)$ terms.

Calculate $S_{n - 1}$

Since $(n - 1)$ is even, we can use the given formula for $S_{n - 1}$ .
Substitute $k = (n - 1)$ into $S_{k} = \frac{k ( k + 1 ) ^{2}}{2}$ .
$S_{n - 1} = \frac{( n - 1 ) (( n - 1 ) + 1 ) ^{2}}{2}$

Simplify $S_{n - 1}$

Simplify the term inside the bracket: $((n - 1) + 1) = n$ .
$S_{n - 1} = \frac{( n - 1 ) n ^{2}}{2}$

Identify the $n$ -th Term

Since $n$ is odd, the $n$ -th term follows the odd-position pattern.
The pattern for odd terms is $T_{k} = k^{2}$ .
Therefore, $T_{n} = n^{2}$ .

Substitute into $S_{n}$ Equation

Recall our split equation: $S_{n} = S_{n - 1} + T_{n}$
Substitute the values we found:
$S_{n} = \frac{( n - 1 ) n ^{2}}{2} + n^{2}$

Factorize the Expression

Both terms have a common factor of $n^{2}$ .
Factor out $n^{2}$ :
$S_{n} = n^{2} (\frac{n - 1}{2} + 1)$

Simplify the Bracket

Take the common denominator inside the bracket:
$S_{n} = n^{2} (\frac{n - 1 + 2}{2})$
$S_{n} = n^{2} (\frac{n + 1}{2})$

Final Result

Multiply the terms to get the final expression.
$S_{n} = \frac{n ^{2} ( n + 1 )}{2}$
This is the sum of the series when $n$ is odd.

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The Sigma Insight: Sum of Special Series

Solution Diagram

Analyzing the Pattern

The given series is

1^{2} + 2 \cdot 2^{2} + 3^{2} + 2 \cdot 4^{2} + 5^{2} + 2 \cdot 6^{2} + \dots

. We can decompose this into two distinct types of terms based on their position index

n

For odd positions (

n = 1, 3, 5, \dots

), the terms follow the pattern

T_{n} = n^{2}

For even positions (

n = 2, 4, 6, \dots

), the terms follow the pattern

T_{n} = 2 \cdot n^{2}

The Even Case

The problem provides a specific formula for the sum of the series when

n

is an even number. This formula accounts for the extra factor of

2

present in every even-positioned term.

The sum for an even number of terms is given by:

S_{n} = \frac{n ( n + 1 ) ^{2}}{2}

The Recursive Strategy for Odd $n$

When

n

is odd, we cannot directly apply the even-sum formula. Instead, we use the property that the term immediately preceding an odd

n

is an even number,

n - 1

We express the total sum

S_{n}

as the sum of the first

n - 1

terms plus the final

n

-th term:

S_{n} = S_{n - 1} + T_{n}

Since

n - 1

is even, we substitute

k = n - 1

into the provided even-sum formula:

S_{n - 1} = \frac{( n - 1 ) (( n - 1 ) + 1 ) ^{2}}{2} = \frac{( n - 1 ) n ^{2}}{2}

The Final Calculation

For an odd position

n

, the term

T_{n}

follows the odd-position rule, which is simply

T_{n} = n^{2}

. We now combine these components into our master equation:

S_{n} = \frac{( n - 1 ) n ^{2}}{2} + n^{2}

To simplify, we factor out the common term

n^{2}

S_{n} = n^{2} (\frac{n - 1}{2} + 1)

By finding a common denominator inside the parentheses, we obtain:

S_{n} = n^{2} (\frac{n - 1 + 2}{2}) = n^{2} (\frac{n + 1}{2})

Thus, the final expression for the sum when

n

is odd is:

$S_{n} = \frac{n ^{2} ( n + 1 )}{2}$

Animated Solution for Mathematics - Sequence and Series: The sum of the first $n$ terms of the series $1^{2} + 2. 2^{2} + 3^{2} + 2. 4^{2} + 5^{2} + 2. 6^{2} + \dots$ is $n (n + 1)^{2} /2$ , when $n$ is even. When $n$ is odd, the sum is \dots.

Visualized Solution

Analyze the Series Pattern

Odd and Even Position Terms

Sum for Even Number of Terms

Strategy for Odd $n$

Splitting the Sum

Calculate $S_{n - 1}$

Simplify $S_{n - 1}$

Identify the $n$ -th Term

Substitute into $S_{n}$ Equation

Factorize the Expression

Simplify the Bracket

Final Result

The Sigma Insight: Sum of Special Series

Analyzing the Pattern

The Even Case

The Recursive Strategy for Odd $n$

The Final Calculation

Similar Questions

The sum of the first $n$ terms of the series $1^{2} + 2. 2^{2} + 3^{2} + 2. 4^{2} + 5^{2} + 2. 6^{2} + \dots$ is $n (n + 1)^{2} /2$ when $n$ is even. When $n$ is odd the sum is

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Visualized Solution

Analyze the Series Pattern

Odd and Even Position Terms

Sum for Even Number of Terms

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

Splitting the Sum

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

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

Identify the .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } n-th Term

Substitute into .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sn​ Equation

Factorize the Expression

Simplify the Bracket

Final Result

The Sigma Insight: Sum of Special Series

Analyzing the Pattern

The Even Case

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

The Final Calculation

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Strategy for Odd $n$

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Simplify $S_{n - 1}$

Identify the $n$ -th Term

Substitute into $S_{n}$ Equation

The Recursive Strategy for Odd $n$

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Sum of the first $n$ terms of the series $\frac{1}{2} + \frac{3}{4} + \frac{7}{8} + \frac{15}{16} + \dots$ is equal to

The sum of series $\frac{1}{2 !} + \frac{1}{4 !} + \frac{1}{6 !} + \dots$ is

The sum of $10$ terms of the series $\frac{3}{1 ^{2} \times 2 ^{2}} + \frac{5}{2 ^{2} \times 3 ^{2}} + \frac{7}{3 ^{2} \times 4 ^{2}} + \dots$ is :

The sum of the first 20 terms of the series $1 + \frac{3}{2} + \frac{7}{4} + \frac{15}{8} + \frac{31}{16} + \dots$ is :

The sum of the series $\frac{1}{1 - 3 \cdot 1 ^{2} + 1 ^{4}} + \frac{2}{1 - 3 \cdot 2 ^{2} + 2 ^{4}} + \frac{3}{1 - 3 \cdot 3 ^{2} + 3 ^{4}} + \dots$ up to 10 terms is

The sum of the series $1 + \frac{1}{4.2 !} + \frac{1}{16.4 !} + \frac{1}{64.6 !} + \dots$ ad inf. is

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