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JEE Mathematics 2025 Question with Solution

The sum 1+3+11+25+45+71+…1 + 3 + 11 + 25 + 45 + 71 + \ldots upto 2020 terms, is equal to

  • A

    72407240

  • B

    71307130

  • C

    69826982

  • D

    81248124

Answer

Correct answer:A

Step-by-step solution

Standard Method

Given: The series is 1+3+11+25+45+71+…1 + 3 + 11 + 25 + 45 + 71 + \ldots and we need the sum of the first 2020 terms.

Find: The value of S=∑n=120TnS = \sum_{n=1}^{20} T_n.

First observe the differences between consecutive terms:

3−1=2,11−3=8,25−11=14,45−25=20,71−45=263 - 1 = 2, \quad 11 - 3 = 8, \quad 25 - 11 = 14, \quad 45 - 25 = 20, \quad 71 - 45 = 26

These first differences form an arithmetic progression with common difference 66.

So the general term can be written as

Tn=1+∑i=1n−1(2+(i−1)×6)T_n = 1 + \sum_{i=1}^{n-1} \left(2 + (i-1) \times 6\right)

Thus,

Tn=1+∑i=1n−1(6i−4)=1+6∑i=1n−1i−4(n−1)T_n = 1 + \sum_{i=1}^{n-1} (6i - 4) = 1 + 6\sum_{i=1}^{n-1} i - 4(n-1)

Using

∑i=1n−1i=(n−1)n2\sum_{i=1}^{n-1} i = \frac{(n-1)n}{2}

we get

Tn=1+3(n−1)n−4(n−1)T_n = 1 + 3(n-1)n - 4(n-1)

So,

Tn=1+(3n2−3n−4n+4)=3n2−7n+5T_n = 1 + (3n^2 - 3n - 4n + 4) = 3n^2 - 7n + 5

Now sum the first 2020 terms:

S=∑n=120Tn=∑n=120(3n2−7n+5)S = \sum_{n=1}^{20} T_n = \sum_{n=1}^{20} (3n^2 - 7n + 5)

Therefore,

S=3∑n=120n2−7∑n=120n+∑n=1205S = 3\sum_{n=1}^{20} n^2 - 7\sum_{n=1}^{20} n + \sum_{n=1}^{20} 5

Using the standard formulae,

∑n=120n2=20×21×416=2870\sum_{n=1}^{20} n^2 = \frac{20 \times 21 \times 41}{6} = 2870 ∑n=120n=20×212=210\sum_{n=1}^{20} n = \frac{20 \times 21}{2} = 210 ∑n=1205=5×20=100\sum_{n=1}^{20} 5 = 5 \times 20 = 100

Substituting these values,

S=3×2870−7×210+100=8610−1470+100=7240S = 3 \times 2870 - 7 \times 210 + 100 = 8610 - 1470 + 100 = 7240

Therefore, the sum of the series up to 2020 terms is 72407240. The correct option is A.

Quadratic Term Assumption

Given: First-order differences of the sequence are in A.P.

Find: The sum of the first 2020 terms.

Since the first differences are in A.P., assume

Tn=an2+bn+cT_n = an^2 + bn + c

Using the first three terms,

T1=1=a+b+cT_1 = 1 = a + b + c T2=3=4a+2b+cT_2 = 3 = 4a + 2b + c T3=11=9a+3b+cT_3 = 11 = 9a + 3b + c

Solving these equations, we get

a=3,b=−7,c=5a = 3, \quad b = -7, \quad c = 5

Hence,

Tn=3n2−7n+5T_n = 3n^2 - 7n + 5

So the required sum is

∑n=120(3n2−7n+5)=320⋅21⋅416−720⋅212+5(20)=7240\sum_{n=1}^{20} (3n^2 - 7n + 5) = 3\frac{20 \cdot 21 \cdot 41}{6} - 7\frac{20 \cdot 21}{2} + 5(20) = 7240

Therefore, the correct option is A.

Common mistakes

  • Assuming the given sequence itself is an A.P. is incorrect because only the first differences form an A.P. Check consecutive differences first, then model TnT_n as a quadratic expression.

  • Writing the nth term incorrectly from the difference pattern is a common error. The differences are 2,8,14,20,26,…2, 8, 14, 20, 26, \ldots with common difference 66, so the incremental term must be handled carefully as 2+(i−1)62 + (i-1)6.

  • Using wrong summation formulae for ∑n\sum n or ∑n2\sum n^2 gives an incorrect final answer. Use ∑n=1Nn=N(N+1)2\sum_{n=1}^{N} n = \frac{N(N+1)}{2} and ∑n=1Nn2=N(N+1)(2N+1)6\sum_{n=1}^{N} n^2 = \frac{N(N+1)(2N+1)}{6}.

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