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JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Sequences and Series

Let

α = 1^{2} + 4^{2} + 8^{2} + 1 3^{2} + 1 9^{2} + 2 6^{2} + \dots

upto 10 terms and

β = \sum_{n = 1}^{10} n^{4}

. If

4 α - β = 55 k + 40

, then

k

is equal to ______.

Answer: 353

Solution

The sequence

1, 4, 8, 13, 19, 26, \dots

has constant second difference

1

, so

t_{n} = a n^{2} + bn + c

. From

t_{1} = 1, t_{2} = 4, t_{3} = 8

:

a = \frac{1}{2}, b = \frac{3}{2}, c = - 1

, i.e.

t_{n} = \frac{n ^{2} + 3 n - 2}{2}

.

4 α = n = 1 \sum 10 (n^{2} + 3 n - 2)^{2} = n = 1 \sum 10 (n^{4} + 6 n^{3} + 5 n^{2} - 12 n + 4) = β + n = 1 \sum 10 (6 n^{3} + 5 n^{2} - 12 n + 4)

since

β = \sum_{n = 1}^{10} n^{4}

. Using

\sum n^{3} = 3025, \sum n^{2} = 385, \sum n = 55

over

n = 1, \dots, 10

:

4 α - β = 6 (3025) + 5 (385) - 12 (55) + 4 (10) = 18150 + 1925 - 660 + 40 = 19455

Setting

19455 = 55 k + 40

gives

k = \frac{19415}{55} = 353

.

JEE Main · 2024 · 30 Jan 2024 · Shift 1 · Integer answer

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