JEE Main 2004

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

Q: Wait, why can't we just use the standard formula for the sum of squares, $\frac{n(n+1)(2n+1)}{6}$?

That formula only applies to the sum of the first $n$ squares ($1^2 + 2^2 + 3^2 + \dots + n^2$). Our series is different! It is $1^2 + 2(2^2) + 3^2 + 2(4^2) + \dots$. Because of that extra factor of 2 on even terms, the standard formula is invalid here. We must derive a new relationship based on the specific structure of this series.

Q: Why is $n-1$ guaranteed to be even when $n$ is odd?

This is a fundamental property of integers! If $n$ is odd, it can be written as $2k+1$ for some integer $k$. Then $n-1 = (2k+1) - 1 = 2k$, which is the definition of an even number. This is why we can safely use the 'even $n$' formula for $S_{n-1}$.

Q: Is there a shortcut to guess the answer?

You could calculate $S_1 = 1^2 = 1$, $S_3 = 1^2 + 2(2^2) + 3^2 = 1 + 8 + 9 = 18$. Then check the options for $n=1$ and $n=3$. For $n=1$, option 2 gives $\frac{1^2(2)}{2} = 1$. For $n=3$, option 2 gives $\frac{3^2(4)}{2} = 18$. It matches! While this is a great way to verify your answer in an exam, the derivation we performed is the rigorous path that ensures your answer is correct for all $n$.

Select Answer:

Progressive Hints

Hint 1

The series is a chameleon. It changes its behavior based on whether the term index

n

is odd or even. Before you jump into formulas, visualize the sequence:

1^{2}, 2 (2^{2}), 3^{2}, 2 (4^{2}), 5^{2}, 2 (6^{2}), \dots

. Notice how the even-positioned terms are 'weighted' by a factor of 2, while the odd-positioned terms are just squares. This parity is the heartbeat of the problem.

Hint 2

You are given a closed-form expression for the sum when

n

is even:

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

. When you are asked for the sum when

n

is odd, you cannot use this formula directly. Instead, use the recursive definition of a series sum:

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

. Since

n

is odd,

n - 1

is even, which allows you to 'borrow' the given formula for

S_{n - 1}

Hint 3

Start by writing

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

. For odd

n

T_{n} = n^{2}

. For

S_{n - 1}

, substitute

(n - 1)

into the given even formula:

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

. Simplify this expression, add

n^{2}

, and factor out the common terms to reach the final result.

Visualized Solution

$A na l y z e t h e S er i es P a tt er n$

Given series: $1^{2} + 2 \cdot 2^{2} + 3^{2} + 2 \cdot 4^{2} + \dots$

$G e n er a l T er m P a tt er n$

If $n$ is odd: $T_{n} = n^{2}$
If $n$ is even: $T_{n} = 2 n^{2}$

$T h e G i v e n I n f or ma t i o n$

Given for even $n$ : $S_{n} = \frac{n ( n + 1 ) ^{2}}{2}$

$T h e L o g i c B r i d g e f or O dd n$

Logic: $S_{n} = S_{n - 1} + T_{n}$

$P a r i t y C h ec k$

Since $n$ is odd, $(n - 1)$ is even.

$F in d in g T_{n}$

For odd $n$ : $T_{n} = n^{2}$

$S u b s t i t u t e in t o t h e E v e n S u m F or m u l a$

Substitute $(n - 1)$ into the even sum formula:
$S_{n - 1} = \frac{( n - 1 ) (( n - 1 ) + 1 ) ^{2}}{2}$

$S im pl i f y S_{n - 1}$

Simplifying the bracket:
$S_{n - 1} = \frac{( n - 1 ) n ^{2}}{2}$

$C o mbin e S_{n - 1} an d T_{n}$

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

$F a c t or i z a t i o n$

Factor out $n^{2}$ :
$S_{n} = n^{2} (\frac{n - 1}{2} + 1)$

$F ina l S im pl i f i c a t i o n$

Simplify the bracket: $\frac{n - 1 + 2}{2} = \frac{n + 1}{2}$

$F ina l R es u l t$

Final Sum for odd $n$ : $S_{n} = \frac{n ^{2} ( n + 1 )}{2}$

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The Sigma Insight: Sum of Special Series

Solution Diagram

Analyzing the Setup

The given series is

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

. This is not a standard arithmetic or geometric progression, so we must identify the underlying pattern.

By observing the terms, we see a clear parity dependence: Odd terms ( $n = 1, 3, 5, \dots$ ): The term is simply $n^{2}$ . Even terms (

n = 2, 4, 6, \dots

): The term is

2 \cdot n^{2}

The Recursive Bridge

The problem provides the sum of the first

n

terms,

S_{n}

, for even

n

as:

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

To find the sum when

n

is odd, we use the recursive relationship:

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

Since

n

is odd,

n - 1

is necessarily even. This allows us to use the provided formula for

S_{n - 1}

and add the

n

-th term,

T_{n} = n^{2}

The Algebraic Dance

We substitute

n - 1

into the given even sum formula:

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

Simplifying the expression inside the parentheses:

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

Now, we combine this with the

n

-th term to find

S_{n}

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

Final Calculation

To simplify the expression, we factor out

n^{2}

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

Combining the terms inside the bracket:

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

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

Thus, for an odd

n

, the sum of the series is: $S_{n} = \frac{n ^{2} ( n + 1 )}{2}$

Common Traps & FAQs

Wait, why can't we just use the standard formula for the sum of squares,

\frac{n ( n + 1 ) ( 2 n + 1 )}{6}

That formula only applies to the sum of the first

n

squares (

1^{2} + 2^{2} + 3^{2} + \dots + n^{2}

). Our series is different! It is

1^{2} + 2 (2^{2}) + 3^{2} + 2 (4^{2}) + \dots

. Because of that extra factor of 2 on even terms, the standard formula is invalid here. We must derive a new relationship based on the specific structure of this series.

Why is

n - 1

guaranteed to be even when

n

is odd?

This is a fundamental property of integers! If

n

is odd, it can be written as

2 k + 1

for some integer

k

. Then

n - 1 = (2 k + 1) - 1 = 2 k

, which is the definition of an even number. This is why we can safely use the 'even

n

' formula for

S_{n - 1}

Is there a shortcut to guess the answer?

You could calculate

S_{1} = 1^{2} = 1

S_{3} = 1^{2} + 2 (2^{2}) + 3^{2} = 1 + 8 + 9 = 18

. Then check the options for

n = 1

and

n = 3

. For

n = 1

, option 2 gives

\frac{1 ^{2} ( 2 )}{2} = 1

. For

n = 3

, option 2 gives

\frac{3 ^{2} ( 4 )}{2} = 18

. It matches! While this is a great way to verify your answer in an exam, the derivation we performed is the rigorous path that ensures your answer is correct for all

n

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

Select Answer:

Visualized Solution

$A na l y z e t h e S er i es P a tt er n$

$G e n er a l T er m P a tt er n$

$T h e G i v e n I n f or ma t i o n$

$T h e L o g i c B r i d g e f or O dd n$

$P a r i t y C h ec k$

$F in d in g T_{n}$

$S u b s t i t u t e in t o t h e E v e n S u m F or m u l a$

$S im pl i f y S_{n - 1}$

$C o mbin e S_{n - 1} an d T_{n}$

$F a c t or i z a t i o n$

$F ina l S im pl i f i c a t i o n$

$F ina l R es u l t$

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Recursive Bridge

The Algebraic Dance

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 \dots.

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 $10$ 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 series $\frac{1}{2 !} + \frac{1}{4 !} + \frac{1}{6 !} + \dots$ is

In the sum of first $n$ terms of an A.P. is $c n^{2}$ , then the sum of squares of these $n$ terms 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 $1 + \frac{1}{4.2 !} + \frac{1}{16.4 !} + \frac{1}{64.6 !} + \dots$ ad inf. is

The sum of the series $\frac{1}{1.2} - \frac{1}{2.3} + \frac{1}{3.4} - \dots$ up to $\infty$ is equal to

If the sum of the first ten terms of the series $(1 \frac{3}{5})^{2} + (2 \frac{2}{5})^{2} + (3 \frac{1}{5})^{2} + 4^{2} + (4 \frac{4}{5})^{2} + \dots$ is $\frac{16}{5} m$ , then $m$ is equal to:

The sum to 20 terms of the series $2 \cdot 2^{2} - 3^{2} + 2 \cdot 4^{2} - 5^{2} + 2 \cdot 6^{2} - \dots$ is equal to __________.

Select Answer:

Visualized Solution

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

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

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

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

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

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

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

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

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

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

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

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

The Sigma Insight: Sum of Special Series

Analyzing the Setup

The Recursive Bridge

The Algebraic Dance

Final Calculation

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the first n terms of the series 12+2.22+32+2.42+52+2.62+… is n(n+1)2/2, when n is even. When n is odd, the sum is \dots.

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Sum of the first n terms of the series 21​+43​+87​+1615​+… is equal to

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of 10 terms of the series 12×223​+22×325​+32×427​+… is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of series 2!1​+4!1​+6!1​+… is

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } In the sum of first n terms of an A.P. is cn2, then the sum of squares of these n terms is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum of the series 1+4.2!1​+16.4!1​+64.6!1​+… ad inf. is

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.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The sum to 20 terms of the series 2⋅22−32+2⋅42−52+2⋅62−… is equal to __________.

$A na l y z e t h e S er i es P a tt er n$

$G e n er a l T er m P a tt er n$

$T h e G i v e n I n f or ma t i o n$

$T h e L o g i c B r i d g e f or O dd n$

$P a r i t y C h ec k$

$F in d in g T_{n}$

$S u b s t i t u t e in t o t h e E v e n S u m F or m u l a$

$S im pl i f y S_{n - 1}$

$C o mbin e S_{n - 1} an d T_{n}$

$F a c t or i z a t i o n$

$F ina l S im pl i f i c a t i o n$

$F ina l R es u l t$

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.

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 $10$ 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 series $\frac{1}{2 !} + \frac{1}{4 !} + \frac{1}{6 !} + \dots$ is

In the sum of first $n$ terms of an A.P. is $c n^{2}$ , then the sum of squares of these $n$ terms 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 $1 + \frac{1}{4.2 !} + \frac{1}{16.4 !} + \frac{1}{64.6 !} + \dots$ ad inf. is

The sum of the series $\frac{1}{1.2} - \frac{1}{2.3} + \frac{1}{3.4} - \dots$ up to $\infty$ is equal to

If the sum of the first ten terms of the series $(1 \frac{3}{5})^{2} + (2 \frac{2}{5})^{2} + (3 \frac{1}{5})^{2} + 4^{2} + (4 \frac{4}{5})^{2} + \dots$ is $\frac{16}{5} m$ , then $m$ is equal to:

The sum to 20 terms of the series $2 \cdot 2^{2} - 3^{2} + 2 \cdot 4^{2} - 5^{2} + 2 \cdot 6^{2} - \dots$ is equal to __________.