If 1-x15-a0-a1 x-a15 x15- then- r -115 r ar -ar-1 is

(1+x)^15 = a_0+a_1+ …… + a_15x^15 So, a_0 = ^15C_0, a_1 = ^15C_1 …… ∑_r=1^15ra_r/a_r 1 = ∑_r=1^15r15!/(15 r)! r!/15!/(15 r+1)!r 1! =r 1̸5̸!/(̸1̸5̸ ̸r̸)̸!̸r̸ ̸×̸ ...

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Byju's Answer

Standard XII

Mathematics

Sum of Coefficients of All Terms

If 1+x15=a0+a...

Question

If $(1 + x)^{15} = a_{0} + a_{1} x + \dots \dots + a_{15} x^{15}$ , then
$\sum_{r = 1}^{15} r \frac{a_{r}}{a_{r} - 1}$ is

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Solution

$(1 + x)^{15} = a_{0} + a_{1} + \dots \dots + a_{15} x^{15}$
So, $a_{0} =^{15} C_{0}, a_{1} =^{15} C_{1} \dots \dots$
$\sum_{r = 1}^{15} r \frac{a_{r}}{a_{r} - 1} = \sum_{r = 1}^{15} r \frac{\frac{15!}{(15 - r)! r!}}{\frac{15!}{(15 - r + 1)! r - 1!}}$
$= r \frac{/ 15!}{/ (15 - r)! r \times / (r - 1)!} \times \frac{(15 - r + 1) . / (15 - r)! / (r - 1)!}{15!}$
$= / r . \frac{15 - r + 1}{/ r} = 16 - r$
$\sum_{r = 1}^{15} (16 - r) = 16 \sum_{r = 1}^{15} 1 - \sum_{r = 1}^{15} r$
$= 16 \times 15 - (1 + 2 + 3 + \dots \dots + 15)$
$= 16 \times 15 - \frac{15 \times 16}{2}$
$= \frac{15 \times 16}{2} = 120$

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If (1+x)15=a0+a1x+……+a15x15, then ∑15r=1rarar−1 is

If $(1 + x)^{15} = a_{0} + a_{1} x + \dots \dots + a_{15} x^{15}$ , then
$\sum_{r = 1}^{15} r \frac{a_{r}}{a_{r} - 1}$ is