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.

Question

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.

Navjyot Kalra · Answer

Hello Student,

Please find the answer to your question

Given that

Runs scored in kth match = k.2^{n + 1 – k}; 1 ≤ k ≤ n

And runs scored in n matches

= n + 1/4 (2^{n + 1} – n – 2)

⇒ 2^{n + 1} $\left [ \sum_{k-1}^{n} \frac{k}{2^k } \right ]$ = n + 1/4 (2^{n + 1} – n – 2)

⇒ 2^{n + 1} [1/2 + 2/2² + 3/2³ + . . . . . . . . . . . +n/2ⁿ]

= n + 1/4 (2^{n + 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/2^{n + 1}

Subtracting the above two, we get

1/2 S = 1/2 + 1/2² + 1/2³ + . . . . . . . . . . . . . . . . . + 1/2ⁿ – n/2^{n + 1}

⇒ 1/2 S = $\frac{\frac{1}{2}(1-\frac{1}{2^n})}{1-\frac{1}{2}}$ - n/2^{n + 1}

⇒ S = 2 [1 – 1/2ⁿ – n/2^{n + 1}]

∴ Equation (i) becomes

2.2^{n + 1} [1 – 1/2ⁿ – n/2^{n + 1}] = n + 1/4 [2^{n + 1} – n – 2]

⇒2 [2^{n + 1} – 2 – n] = n + 1/4 [2^{n + 1} – 2 – n]

⇒ n + 1/4 = 2 ⇒ n = 7.

Thanks
navjot kalra
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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 (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.

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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.

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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.