Rational Numbers | IT

Question 28

Context: Look at the following pattern of consecutive odd numbers: $1 = 1 = 1^3$ $3 + 5 = 8 = 2^3$ $7 + 9 + 11 = 27 = 3^3$ $13 + 15 + 17 + 19 = 64 = 4^3$ $21 + 23 + 25 + 27 + 29 = 125 = 5^3$ $31 + 33 + 35 + 37 + 39 + 41 = 216 = 6^3$

Later in this series, we get the following set of consecutive numbers: $91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109$

Q. Can you tell what this sum is without doing the calculation?

Solution

We can find the row number of the given series and then use the pattern to calculate its sum.

Step 1 — Find the number of terms Let us first find how many numbers are in the given series. The series is $91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109$ . These are consecutive odd numbers. Let $a_1$ be the first term. So, $a_1 = 91$ . Let $a_k$ be the last term. So, $a_k = 109$ . The common difference between consecutive odd numbers is $d = 2$ . We use the formula for the $k$ -th term of an arithmetic progression. This formula is $a_k = a_1 + (k-1)d$ . We substitute the values we know into the formula. $109 = 91 + (k-1) \times 2$ $109 - 91 = (k-1) \times 2$ $18 = (k-1) \times 2$ $\frac{18}{2} = k-1$ $9 = k-1$ $k = 9+1$

$\boxed{\text{Number of terms} = 10}$

Step 2 — Find the row number We need to find which row in the pattern has $91$ as its first number. Let $F_n$ be the first term of the $n$ -th row. From the given pattern, we observe the first terms: $F_1 = 1$ $F_2 = 3$ $F_3 = 7$ $F_4 = 13$ $F_5 = 21$ $F_6 = 31$ Let us look at the difference between consecutive first terms. $F_2 - F_1 = 2$ . $F_3 - F_2 = 4$ . $F_4 - F_3 = 6$ . $F_5 - F_4 = 8$ . $F_6 - F_5 = 10$ . The difference for the $n$ -th row is $2(n-1)$ . So, $F_n = F_{n-1} + 2(n-1)$ . We can write a general formula for $F_n$ . $F_n = F_1 + \text{sum of differences up to } (n-1)\text{-th term}$ . $F_n = 1 + 2 + 4 + 6 + \dots + 2(n-1)$ . $F_n = 1 + 2 \times (1 + 2 + 3 + \dots + (n-1))$ . The sum of the first $(n-1)$ natural numbers is $\frac{(n-1)n}{2}$ . So, $F_n = 1 + 2 \times \frac{(n-1)n}{2}$ . $F_n = 1 + n(n-1)$ $F_n = n^2 - n + 1$ The first term of our series is $91$ . So, we set $F_n = 91$ . $n^2 - n + 1 = 91$ $n^2 - n - 90 = 0$ We need to solve this quadratic equation for $n$ . We can factor the equation. $(n-10)(n+9) = 0$ This gives two possible values for $n$ : $n=10$ or $n=-9$ . Since $n$ must be a positive row number, we choose the positive value.

$\boxed{\text{Row number} = 10}$

Step 3 — Calculate the sum We have found that the given series is the 10th row in the pattern. The pattern shows that the sum of the $n$ -th row is $n^3$ . For example, the 1st row sum is $1^3 = 1$ . The 2nd row sum is $2^3 = 8$ . The 3rd row sum is $3^3 = 27$ . Since our series is the 10th row, we use $n=10$ . The sum of the series will be $10^3$ . $10^3 = 10 \times 10 \times 10$

$\boxed{\text{Sum} = 1000}$

Answer

(i) The sum is 1000.

More questions in IT

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Q11

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Q12

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Q14

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Q15

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Q16

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Q17

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Q18

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Q19

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Q20

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Q21

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Q22

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Q23

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Q24

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Q25

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Q26

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Q27

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Q28

Later in this series, we get the following set of consecutive numbers: $91 + 93 + 95 + 97 + 99 + 101 + 103 + 105 + 107 + 109$

Q. Can you tell what this sum is without doing the calculation?

Q29

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Q30

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