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If total number of runs scored in n matches is (n + 1)/4 (2 n + 1 – n – 2) where n > 1, and the runs scored in the k th match are given by k. 2 n + 1 – k , where 1 ≤ k ≤ n. Find n.

If total number of runs scored in n matches is (n + 1)/4 (2n + 1 – n – 2) where n > 1, and the runs scored in the k th match are given by k. 2n + 1 – k, where 1 ≤ k ≤ n. Find n.

Grade:upto college level

1 Answers

Navjyot Kalra
askIITians Faculty 654 Points
7 years ago
Hello Student,
Please find the answer to your question
Given that
Runs scored in kth match = k.2n + 1 – k; 1 ≤ k ≤ n
And runs scored in n matches
= n + 1/4 (2n + 1 – n – 2)
232-732_121.png
⇒ 2n + 1 \left [ \sum_{k-1}^{n} \frac{k}{2^k } \right ]= n + 1/4 (2n + 1 – n – 2)
⇒ 2n + 1 [1/2 + 2/22 + 3/23 + . . . . . . . . . . . +n/2n]
= n + 1/4 (2n + 1 – n – 2) . . . . . . . . . . . . . . . . . . . . (i)
Let S = 1/2 + 2/22 + 3/23 + . . . . . . . . . . . . . . . n/2n
1/2 S = 1/22 + 2/23 + . . . . . . . . . . . . . . . . . + n – 1/2n + n/2n + 1
Subtracting the above two, we get
1/2 S = 1/2 + 1/22 + 1/23 + . . . . . . . . . . . . . . . . . + 1/2n – n/2n + 1
⇒ 1/2 S = \frac{\frac{1}{2}(1-\frac{1}{2^n})}{1-\frac{1}{2}}- n/2n + 1
⇒ S = 2 [1 – 1/2n – n/2n + 1]
∴ Equation (i) becomes
2.2n + 1 [1 – 1/2n – n/2n + 1] = n + 1/4 [2n + 1 – n – 2]
⇒2 [2n + 1 – 2 – n] = n + 1/4 [2n + 1 – 2 – n]
⇒ n + 1/4 = 2 ⇒ n = 7.

Thanks
navjot kalra
askIITians Faculty

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