Sum of the series 3C1 - 4C2-5C3-6C4- upto n terms is where Cr- nCr

Let S=3C_1 4C_2+5C_3 7C_5+··· upto n terms =∑_r=1^n( 1)^r 1 (r+2)C_r nbsp; =∑_r=1^n( 1)^r 1· r· ^nC_r+2∑_r=1^n( 1)^r 1· ^nC_r nbsp; =∑_r=1^n( 1)^r 1· n· ^ ...

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Standard XI

Mathematics

Sum of Binomial Coefficients with Alternate Signs

Sum of the se...

Question

Sum of the series $3 C_{1} - 4 C_{2} + 5 C_{3} - 6 C_{4} + \dots$ upto $n$ terms is (where $C_{r} =^{n} C_{r}$ )

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Solution

The correct option is B $2$
Let $S = 3 C_{1} - 4 C_{2} + 5 C_{3} - 7 C_{5} + \cdot \cdot \cdot$ upto $n$ terms
$= n \sum r = 1 (- 1)^{r - 1} (r + 2) C_{r}$
$= n \sum r = 1 (- 1)^{r - 1} \cdot r \cdot^{n} C_{r} + 2 n \sum r = 1 (- 1)^{r - 1} \cdot^{n} C_{r}$
$= n \sum r = 1 (- 1)^{r - 1} \cdot n \cdot^{n - 1} C_{r - 1} - 2 n \sum r = 1 (- 1)^{r} \cdot^{n} C_{r}$
$= n \cdot 0 - 2 n \sum r = 0^{n} C_{r} (- 1)^{r} + 2^{n} C_{0} (∵ n \sum r = 0^{n} C_{r} (a)^{r} = {(1 + a)}^{n})$
$= 2$

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Sum of Binomial Coefficients with Alternate Signs

Standard XI Mathematics

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Sum of the series 3C1−4C2+5C3−6C4+⋯ upto n terms is (where Cr= nCr)

The correct option is B 2Let S=3C1−4C2+5C3−7C5+⋅⋅⋅ upto n terms =n∑r=1(−1)r−1 (r+2)Cr =n∑r=1(−1)r−1⋅r⋅nCr+2n∑r=1(−1)r−1⋅nCr =n∑r=1(−1)r−1⋅n⋅n−1Cr−1−2n∑r=1(−1)r⋅nCr =n⋅0−2n∑r=0nCr(−1)r+2 nC0 (∵n∑r=0nCr(a)r=(1+a)n) =2

Sum of the series $3 C_{1} - 4 C_{2} + 5 C_{3} - 6 C_{4} + \dots$ upto $n$ terms is (where $C_{r} =^{n} C_{r}$ )