Answer:
The 12th partial sum of the given series is:
-276
Step-by-step explanation:
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
