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lilavasa [31]
3 years ago
13

Find the 12th partial sum of ∞∑  (-2i-10)i-1

Mathematics
2 answers:
zvonat [6]3 years ago
8 0
<h2>Answer:</h2>

The 12th partial sum of the given series is:

                                -276

<h2>Step-by-step explanation:</h2>

We are asked to find the 12th partial sum of the series which is given by:

             \sum_{i=1}^{\infty}(-2i-10)

i.e. we have to find the sum of first 12 terms.

i.e.

\sum_{i=1}^{12}(-2i-10)

which could also be simplified by the method:

\sum_{i=1}^{12}(-2i-10)=-2\sum_{i=1}^{12}i-10\sum_{i=1}^{12}1

which is further given by:

\sum_{i=1}^{12}(-2i-10)=-2\times (1+2+3+4+5+6+7+8+9+10+11+12)-10\times (1+1+1+1+1+1+1+1+1+1+1+1)

know we know that:

1+2+3+.....+n=\dfrac{n(n+1)}{2}

i.e.

1+2+3+....+12=\dfrac{12\times 13}{2}\\\\1+2+3+....+12=78

Hence, we get:

\sum_{i=1}^{12}(-2i-10)=-2\times 78-10\times 12\\\\[tex]\sum_{i=1}^{12}(-2i-10)=-156-120\\\\[tex]\sum_{i=1}^{12}(-2i-10)=-276

           Hence, the answer is:  -276

telo118 [61]3 years ago
4 0
Hello,


$\sum_{i=1}^{12}(-2i-10)=-10\sum_{i=1}^{12}1-2\sum_{i=1}^{12}i$

=-120-2*13*12/2=-120-156=-276

 
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