JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Sequences and Series

Let

a_{1} = b_{1} = 1, a_{n} = a_{n - 1} + 2

and

b_{n} = a_{n} + b_{n - 1}

for every natural number

n \geq 2

. Then

\sum_{n = 1}^{15} a_{n} \cdot b_{n}

is equal to ______.

Answer: 27560

Solution

First, find the general terms for

a_{n}

and

b_{n}

.

For

a_{n}

: Since

a_{1} = 1

and

a_{n} = a_{n - 1} + 2

, this is an arithmetic sequence with first term 1 and common difference 2. Thus

a_{n} = 2 n - 1

.

For

b_{n}

: We have

b_{1} = 1

b_{2} = a_{2} + b_{1} = 3 + 1 = 4

b_{3} = a_{3} + b_{2} = 5 + 4 = 9

b_{4} = a_{4} + b_{3} = 7 + 9 = 16

. By pattern recognition,

b_{n} = n^{2}

.

Now compute:

n = 1 \sum 15 a_{n} b_{n} = n = 1 \sum 15 (2 n - 1) n^{2} = n = 1 \sum 15 (2 n^{3} - n^{2}) = 2 n = 1 \sum 15 n^{3} - n = 1 \sum 15 n^{2}

Using standard formulas:

\sum_{n = 1}^{15} n^{3} = \frac{1 5 ^{2} \cdot 1 6 ^{2}}{4} = \frac{225 \cdot 256}{4} = 14400

and

\sum_{n = 1}^{15} n^{2} = \frac{15 \cdot 16 \cdot 31}{6} = 1240

.

Therefore:

2 (14400) - 1240 = 28800 - 1240 = 27560

JEE Main · 2022 · 25 Jul 2022 · Shift 1 · Integer answer