Question

Which number needs to be the last term of the sequence 1, -2,3,-4,5,... so that
the sum would be 2011?​

Answer 1

Answer:

4021

Step-by-step explanation:

1+(-2)+3+(-4).....

(1-2)+(3-4)+5....

so even number gets deducted from its previous odd number and every such group of 2 numbers (odd plus even) gives -1 as a value.

so

we will have

-1-1-1-1+odd number...

if we get up to 4021. previous groups summations will be -2010 (as every 2 numbers give the value of -1)

so last will be

-2010+4021= 2011

Answer 2

Given sequence $1,-2,3,-4,....$

If we look closely and write the series as, $(1-2),(3-4),(5-6), (7-8),...$

The sum of each bracket is equal to $(-1)$

But, if we write the series as,

$(1-2)+3\rightarrow -1+3\rightarrow 2\\(1-2)+(3-4)+5\rightarrow -1-1+5\rightarrow -2+5\rightarrow 3\\(1-2)+(3-4)+(5-6)+7\rightarrow -1-1-1+7\rightarrow -3+7\rightarrow 4$

We can see that the sum is $-[\frac{x}{2}]+x$ and x is an odd number.

Also, the sum is always $\left |-[\frac{x}{2}]}\right |+1$ and x is an odd number.

Also, x will be the number of terms in the series.

So sum = 2011

$\therefore \left |-[\frac{x}{2}]}\right |+1=2011\\\left |-[\frac{x}{2}]}\right |=2010\\\therefore x=4021$

#SPJ2