Find the sum of series $20^2 -19^2 + 18^2 - 17^2 +16^2 -15^2 +...+ 2^2 -1^2$?

Question

Find the sum of series $20^2 -19^2 + 18^2 - 17^2 +16^2 -15^2 +...+ 2^2 -1^2$

The following working is found on Socratic, but I don’t understand it.

Is there another method of solving the question or can someone explain the working?

Thank you

Answer 1

$n^2-(n-1)^2 = n^2-(n^2-2n+1) = 2n-1 = n+(n-1)$

so

$20^2-19^2=20+19\\18^2-17^2 = 18+17\\16^2-15^2=16+15$

etc. Does this help you ?

Answer 2

$$\sum_{k=1}^{10}\left(2k\right)^{2}-\sum_{k=0}^{9}\left(2k+1\right)^{2}=4\sum_{k=1}^{10}k^{2}-\left(4\sum_{k=0}^{9}\left(k\right)^{2}+4\sum_{k=0}^{9}\left(k\right)+\sum_{k=0}^{9}1\right)$$$$=4\cdot\frac{10\left(11\right)\left(21\right)}{6}-\left(4\cdot\frac{9\left(10\right)\left(19\right)}{6}+4\cdot\frac{9\left(10\right)}{2}+10\right)$$$$=210$$

This is true assuming $0/0=1$

Also for $\sum_{k=0}^{9}1$ equivalently we have :$$\sum_{k=0}^{9}1=\sum_{k=0}^{9}\frac{k}{k}=10$$

Another well-known formulas I've used:

$$\sum_{k=0}^{n}k=\frac{n\left(n+1\right)}{2},\sum_{k=0}^{n}k^{2}=\frac{n\left(n+1\right)\left(2n+1\right)}{6}$$

Answer 3

$(20^2-19^2)+(18^2-17^2)+$

$......(4^2-3^2)+(2^2-1^1)=$

$(20+19)+(18+17)+ (16+15)$

$+.......+(4+3)+(2+1)=$

$39 +35+31+.......+3$ ($10$ terms).

In reverse order:

$3+7+11+.........+39$

Add:

$42+42+...........+42$ ($10$ terms)

$2$ Sum $=420$;

Sum $=210$.

Used: $m^2-n^2=(m+n)(m-n).$

Answer 4

If you are fine with the formula $1^2+2^2+ \cdots + n^2 =\frac 16 n(n+1)(2n+1)$, then you can calculate the sum almost directly without any splitting of the summands as follows:

• Let's call $S_{20} = 1^2+2^2+ \cdots + 20^2$.
• Then, summing only the even powers gives $2^2 + 4^2 + \cdots + 20^2 = 4(1^2 + 2^2 + \cdots + 10^2) = 4S_{10}$.
• So, for the odd ones it follows $1^2 + 3^2 + \cdots + 19^2 = S_{20}-4S_{10}$.
• Hence, even ones minus odd ones is

$$20^2 - 19^2 + \cdots + 2^2 - 1^1 =4S_{10} - (S_{20}-4S_{10})= 8S_{10}-S_{20}$$ $$\stackrel{formula}{=}\frac 16(8\cdot10\cdot 11\cdot 21 - 20\cdot 21\cdot 41)=210$$

Answer 5

Computing directly via generating functions: $$[x^{20}]\frac1{1-x}\left(x \frac d{dx}\right)^2\frac1{1+x} = [x^{20}]{-x\over(1+x)^3} = [x^{19}]\frac{-1}{(1+x)^3} = (-1)^{20}\binom{3+19-1}{19} = 210.$$ However, the more clever solution is to notice that by pairing up adjacent terms you can reduce the sum to the sum of the whole numbers from $1$ through $20$.