The sum 1 + 3 + 11 + 25 + 45 + 71 + ... upto 20 terms, is equal to

The correct answer is Option 3. Full mathematical solution steps and hints are accessible in the interactive StemCET workspace.

Maths Q7 – JEE-MAIN 2025

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Home Maths Sequences and SeriesQuestion #408

The sum

1 + 3 + 11 + 25 + 45 + 71 + \dots

upto 20 terms, is equal to

Clue 1 of 3: Identify the type of sequence

Calculate the differences between consecutive terms to determine if it's an arithmetic progression of higher order.

🥷

Ninja Strategy•Divisibility by 10

Derive the general sum formula $S_{N} = N (N^{2} - 2 N + 2)$ . Since $N = 20$ (a multiple of 10), the sum $S_{20}$ must also be a multiple of 10, which means its last digit must be 0. This eliminates options not ending in 0.

🧠 Full Solution Path

Step 1: Determine the General Term (

T_{n}

)✦ Active

Let the given series be $S = T_{1} + T_{2} + T_{3} + \dots + T_{20}$ . The terms are $1, 3, 11, 25, 45, 71, \dots$ . Calculate the differences between consecutive terms:

D_{1} : 3 - 1 = 2, 11 - 3 = 8, 25 - 11 = 14, 45 - 25 = 20, 71 - 45 = 26, \dots

Now, calculate the differences of the first differences:

D_{2} : 8 - 2 = 6, 14 - 8 = 6, 20 - 14 = 6, 26 - 20 = 6, \dots

Since the second differences are constant ( $6$ ), the general term $T_{n}$ is a quadratic polynomial of the form $A n^{2} + B n + C$ . Using the relations for coefficients:

2 A = D_{2} ⟹ 2 A = 6 ⟹ A = 3

3 A + B = D_{1} (1) ⟹ 3 (3) + B = 2 ⟹ 9 + B = 2 ⟹ B = - 7

A + B + C = T_{1} ⟹ 3 - 7 + C = 1 ⟹ - 4 + C = 1 ⟹ C = 5

Thus, the general term is $T_{n} = 3 n^{2} - 7 n + 5$ .

Step 2: Calculate the Sum of the Series (

S_{N}

)○ Expand

The sum of the first $N$ terms is $S_{N} = \sum_{n = 1}^{N} T_{n} = \sum_{n = 1}^{N} (3 n^{2} - 7 n + 5)$ . We need to find $S_{20}$ .

S_{N} = 3 \sum_{n = 1}^{N} n^{2} - 7 \sum_{n = 1}^{N} n + \sum_{n = 1}^{N} 5

Using the standard summation formulas: $\sum_{n = 1}^{N} n = \frac{N (N + 1)}{2}$ and $\sum_{n = 1}^{N} n^{2} = \frac{N (N + 1) (2 N + 1)}{6}$ .

Substitute $N = 20$ :

\sum_{n = 1}^{20} n = \frac{20 (21)}{2} = 210

\sum_{n = 1}^{20} n^{2} = \frac{20 (21) (41)}{6} = 10 (7) (41) = 2870

💡 Teacher's Secret Hint

Remember to correctly apply the summation formulas for $n$ and $n^{2}$ .

Step 3: Compute the Final Sum○ Expand

Substitute the summation values into the expression for $S_{20}$ :

S_{20} = 3 (2870) - 7 (210) + 5 (20)

S_{20} = 8610 - 1470 + 100

S_{20} = 7140 + 100

S_{20} = 7240

Alternatively, a simplified formula for $S_{N}$ can be derived: $S_{N} = N (N^{2} - 2 N + 2)$ . For $N = 20$ :

S_{20} = 20 (20^{2} - 2 (20) + 2) = 20 (400 - 40 + 2) = 20 (362) = 7240

💡 Teacher's Secret Hint

Double-check your arithmetic to avoid calculation errors.

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Maths Question 7 – JEE-MAIN 2025

Clue 1 of 3: Identify the type of sequence

🧠 Full Solution Path