Special Series in Sequences and Series

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$The positive integer ' n' for which 2 . 2^{2} + 3 . 2^{3} + 4 . 2^{4} + . . + n . 2^{n} = 2^{n + 10} is$

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Solution

$\begin{array}{l} 2^{n + 10} = {2.2}^{2} + {3.2}^{3} + {4.2}^{4} + . . + n \cdot 2^{n} \\ \frac{2 (2^{n + 10}) = {2.2}^{3} + {3.2}^{4} + . . + (n - 1) 2^{n} + n \cdot 2^{n + 1}}{- 2^{n + 10} = 2 \cdot 2^{2} + 2^{3} + 2^{4} + . . + 2^{n} - n \cdot 2^{n + 1}} \\ = 8 + 8 (2^{2 n - 2} - 1) - n \cdot 2^{n + 1} \\ = 8 + 2^{n + 1} - 8 - n \cdot 2^{n + 1} = 2^{n + 1} - n \cdot 2^{n + 1} \\ n = 513 \end{array}$

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Special Series in Sequences and Series

Remember concepts with our Masterclasses.

The positive integer ' n ' for which 2.22+3.23+4.24+..+n.2n=2n+10 is

2n+10=2.22+3.23+4.24+..+n⋅2n22n+10=2.23+3.24+..+(n−1)2n+n⋅2n+1−2n+10=2⋅22+23+24+..+2n−n⋅2n+1=8+822n−2−1−n⋅2n+1=8+2n+1−8−n⋅2n+1=2n+1−n⋅2n+1n=513

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$The positive integer ' n' for which 2 . 2^{2} + 3 . 2^{3} + 4 . 2^{4} + . . + n . 2^{n} = 2^{n + 10} is$