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NVAMediumJEE 2024Sum of Series

JEE Mathematics 2024 Question with Solution

Let α\alpha = 12+42+82+132+192+262+⋯1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + \cdots up to 1010 terms and β=∑n=110n4\beta = \sum_{n=1}^{10} n^4. If 4α−β=55k+404\alpha - \beta = 55k + 40, then kk is equal to:

Answer

Correct answer:353

Step-by-step solution

Standard Method

Given:

  • α=12+42+82+132+192+262+⋯\alpha = 1^2 + 4^2 + 8^2 + 13^2 + 19^2 + 26^2 + \cdots up to 1010 terms
  • β=∑n=110n4\beta = \sum_{n=1}^{10} n^4
  • 4α−β=55k+404\alpha - \beta = 55k + 40

Find: kk

First identify the sequence whose squares are being added in α\alpha: 1,4,8,13,19,26,…1, 4, 8, 13, 19, 26, \ldots

The successive differences are 3,4,5,6,7,…3,4,5,6,7,\ldots, so the sequence is quadratic. Let its nn-th term be

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

Using

T1=1,T2=4,T3=8T_1 = 1, \quad T_2 = 4, \quad T_3 = 8

we get

a+b+c=1a + b + c = 1 4a+2b+c=44a + 2b + c = 4 9a+3b+c=89a + 3b + c = 8

Solving,

a=12,b=32,c=−1a = \frac{1}{2}, \quad b = \frac{3}{2}, \quad c = -1

Hence

Tn=12n2+32n−1T_n = \frac{1}{2}n^2 + \frac{3}{2}n - 1

Therefore,

α=∑n=110(12n2+32n−1)2\alpha = \sum_{n=1}^{10} \left(\frac{1}{2}n^2 + \frac{3}{2}n - 1\right)^2

So,

4α=∑n=110(n2+3n−2)24\alpha = \sum_{n=1}^{10} (n^2 + 3n - 2)^2

From the detailed computation of the terms:

a1=1, a2=4, a3=8, a4=13, a5=19, a6=26, a7=34, a8=43, a9=53, a10=64a_1=1,\ a_2=4,\ a_3=8,\ a_4=13,\ a_5=19,\ a_6=26,\ a_7=34,\ a_8=43,\ a_9=53,\ a_{10}=64

Hence,

α=1+16+64+169+361+676+1156+1849+2809+4096=11197\alpha = 1+16+64+169+361+676+1156+1849+2809+4096 = 11197

Now compute

β=∑n=110n4=14+24+34+⋯+104=25333\beta = \sum_{n=1}^{10} n^4 = 1^4+2^4+3^4+\cdots+10^4 = 25333

Substitute into the given relation:

4α−β=4⋅11197−25333=44788−25333=194554\alpha - \beta = 4 \cdot 11197 - 25333 = 44788 - 25333 = 19455

Thus,

19455=55k+4019455 = 55k + 40 55k=1941555k = 19415 k=1941555=353k = \frac{19415}{55} = 353

Therefore, the value of kk is 353353.

Pattern Recognition Method

Given: the base sequence in α\alpha is 1,4,8,13,19,26,…1,4,8,13,19,26,\ldots

Find: kk

Observe the pattern of differences:

3,4,5,6,7,…3,4,5,6,7,\ldots

This shows that each term can be written directly as

an=1+∑r=1n−1(r+2)a_n = 1 + \sum_{r=1}^{n-1} (r+2)

which gives

an=n(n+1)2+1=n2+n+22a_n = \frac{n(n+1)}{2} + 1 = \frac{n^2+n+2}{2}

Now evaluate the first 1010 terms:

1,4,8,13,19,26,34,43,53,641,4,8,13,19,26,34,43,53,64

So

α=12+42+82+132+192+262+342+432+532+642=11197\alpha = 1^2+4^2+8^2+13^2+19^2+26^2+34^2+43^2+53^2+64^2 = 11197

Also,

β=∑n=110n4=25333\beta = \sum_{n=1}^{10} n^4 = 25333

Hence,

4α−β=4(11197)−25333=194554\alpha - \beta = 4(11197) - 25333 = 19455

Using

19455=55k+4019455 = 55k + 40

we get

k=19455−4055=353k = \frac{19455-40}{55} = 353

Therefore, the value of kk is 353353.

Common mistakes

  • A common mistake is to assume 1,4,8,13,19,26,…1,4,8,13,19,26,\ldots is an arithmetic progression. That is wrong because the first differences are not constant. Instead, check the successive differences and recognize the sequence as quadratic.

  • Some students compute α\alpha as 1+4+8+13+⋯1+4+8+13+\cdots instead of 12+42+82+132+⋯1^2+4^2+8^2+13^2+\cdots. This changes the problem completely. The terms in α\alpha are the squares of the sequence terms, so square each term before summing.

  • Another mistake is to use an incorrect formula for β\beta. Here β=∑n=110n4\beta = \sum_{n=1}^{10} n^4, so the sum of fourth powers is required, not the sum of squares or cubes. Write the powers carefully before evaluating.

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