If the sum of the series 1-2-2-3-22-4-23-100-299 is knn-1-m- then find k-m-

S=1+2·2+3· 2 ^ 2 +4· 2 ^ 3 +…+100· 2 ^ 99 ⇒ 2S=2+2· 2 ^ 2 +3· 2 ^ 3 +…+100· 2 ^ 100 ⇒ (1 2)S=1+2+ 2 ^ 2 + 2 ^ 3 +·+ 2 ^ 99 100· 2 ^ 100 ⇒ ...

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Standard XII

Mathematics

Sum of Product of Binomial Coefficients

If the sum of...

Question

If the sum of the series $1 + 2 \cdot 2 + 3 \cdot 2^{2} + 4 \cdot 2^{3} + \dots + 100 \cdot 2^{99}$ is $k^{n} (n - 1) + m$ , then find $k + m$ .

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Solution

$S = 1 + 2 \cdot 2 + 3 \cdot 2^{2} + 4 \cdot 2^{3} + \dots + 100 \cdot 2^{99}$
$\Rightarrow 2 S = 2 + 2 \cdot 2^{2} + 3 \cdot 2^{3} + \dots + 100 \cdot 2^{100}$
$\Rightarrow (1 - 2) S = 1 + 2 + 2^{2} + 2^{3} + \cdot + 2^{99} - 100 \cdot 2^{100}$
$\Rightarrow - S = \frac{2^{100} - 1}{2 - 1} - 100 \cdot 2^{100} = 2^{100} - 1 - 100 \cdot 2^{100}$
$\Rightarrow - S = - {99 \cdot 2}^{100} - 1 \Rightarrow S = (100 - 1) \cdot 2^{100} + 1 ∴ k + m = 2 + 1 = 3$

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If the sum of the series 1+2⋅2+3⋅22+4⋅23+⋯+100⋅299 is kn(n−1)+m, then find k+m.

S=1+2⋅2+3⋅22+4⋅23+⋯+100⋅299⇒2S=2+2⋅22+3⋅23+⋯+100⋅2100⇒(1−2)S=1+2+22+23+⋅+299−100⋅2100⇒−S=2100−12−1−100⋅2100=2100−1−100⋅2100⇒−S=−99⋅2100−1⇒S=(100−1)⋅2100+1∴k+m=2+1=3

If the sum of the series $1 + 2 \cdot 2 + 3 \cdot 2^{2} + 4 \cdot 2^{3} + \dots + 100 \cdot 2^{99}$ is $k^{n} (n - 1) + m$ , then find $k + m$ .