Modified Pyramids 2

$1=0^3 + 1^3$

Number of terms in LHS = 1.

Sum of numbers in RHS $=0+1$ .

$2+3+4=1^3+2^3$

Number of terms in LHS = 3.

Sum of numbers in RHS $=1+2=3$ .

$5+6+7+8+9=2^3+3^3$

Number of terms in LHS = 5.

Sum of numbers in RHS $=2+3=5$ .

$10+11+12+13+14+15+16=3^3+4^3$

Number of terms in LHS = 7.

Sum of numbers in RHS $=3+4=7$ .

$\dots \\ \dots$ .

$9802+9803 + \dots + 9999 + 10000 = A^3 + (A+1)^3$

Number of terms in LHS $=10000-9802+1 = 198+1=199$ .

Thus we can write equation for sum of numbers:

$A+A+1=199 \\ \Rightarrow 2A+1=199 \\\Rightarrow 2A=198 \\ \Rightarrow \Large\boxed{A=99}$

we can easily get value of A by using induction

by observing given equalities we derived a general formula for them which is 》

(k^2-(2(k-1))+........(k^2 -2)+(k^2-1)+k^2=(k-1)^3+k^3

but we dont know if our general formula is true for all values of k. now according to mathematical induction any general formula in variable k is true for all values of k

if it is true for values,

¤k=1.

¤k=k+1.

and after putting these values for k in general formula.

we found that both are true so this is true.

now by comparing our general formula with formula given in question. we cam easily get value of variable A,which is equal to 99.

by comparing we get.

10000=(A+1)^2.

A^2+2A-9999=0.

A^2+101A-99A-9999=0.

(A+101)(A-99)=0.

A=-101,99.

reject negative value of A(it should be given in question that A is a natural number otherwise both will be true) so A=99.

we see that this is very easy question only deriving general formula is a little hard,but it is fun to derive general formula its like solving a puzzle . :)

If we had an arithmetic series like $1+2+3+...+n$ then the sum of all elements in the series would be $\frac{n}{2}*(1+n)$ . $1+2+3+...+10000 =\frac{10000}{2}*(1+10000)$ $1+2+3+...+9801 = \frac{9801}{2}*(1+9801)$ By subtracting the second from the first, we wil be having $9802+9803+...+9999+10000 = (\frac{10000}{2}*(1+10000))-(\frac{9801}{2}*(1+9801))$ At the end : $A^3+ (A+1)^3 = (\frac{10000}{2}*(1+10000))-(\frac{9801}{2}*(1+9801))$

$A^3+ (A+1)^3 = (5000*10001)-(9801*4901)$

$A^3+ (A+1)^3 = (50005000)-(40095891)$

$A^3+ (A+1)^3 = 1970299$

$\boxed{A = 99}$

Some may not see this but, i can show a pattern.

On the first line, we can see that 1 is obviously 1^3, well on the second we can see 2+3+4=1^3+2^3, what i want to point out is the relationship of the last terms of each side of each equation, 4=2^2, 2^3; next line: 9=3^2 , 3^3; then the next 16=4^2, 4^3

So for finding A, take the square root of 10000 which is 100- this is A+1

A+1=100, A=99

Add all equations to get:

$1+ 2 + 3 + \cdots + 9999 + 10000 = 2 \cdot (1^{3} + 2^{3} + 3^{3} + \cdots + A^{3}) + (A + 1)^{3}$

$\sum_{n=1}^{10000} n = 2 \cdot ( \sum_{n=1}^{A} n^{3}) + (A + 1)^{3}$

$\frac{10000 \cdot (10000 + 1)}{2} = 2 \cdot ( \frac{A^{2} \cdot (A+1)^{2}}{4}) + (A + 1)^{3}$

$10000 \cdot (10000 + 1) = A^{2} \cdot (A+1)^{2} + 2 \cdot (A + 1)^{3}$

$10000 \cdot (10000 + 1) = (A + 1)^{2} \cdot (A^{2} + 2 \cdot (A + 1))$

$100^{2} \cdot (10000 + 1) = (A + 1)^{2} \cdot (A^{2} + 2 \cdot (A + 1))$

$(99 + 1)^{2} \cdot (10000 + 1) = (A + 1)^{2} \cdot (A^{2} + 2 \cdot (A + 1))$

At this point, we might conjecture that $A = 99$ . To prove it, we just need to show that the second part of our product holds, if $A = 99$ , i.e. we need to prove the following equality:

$10000 + 1 = 99^{2} + 2 \cdot (99 + 1)$

$10001 = 99 \cdot 99 + 2 \cdot 100$

$10001 = (100 - 1) \cdot 99 + 2 \cdot 100$

$10001 = 99 \cdot 100 + 2 \cdot 100 - 99$

$10001 = (99 + 2) \cdot 100 - 99$

$10001 = (101) \cdot 100 - 99$

$10001 = 10100 - 99$

$10100 = 10100$

So, the equality holds and we have proven that $\boxed{A = 99}$ .

how I did the problem is that I observed that the 1st term in RHS squared + 1 is equal to LHS first term.

So, A^2+1=9802. so, A = 99.

Let the number of terms on the LHS of $k^{th}$ row be $n_k$ . The we have:

$\begin{array} {rlll} 1 & = 0^3 + \color{#3D99F6}{1}^3 & \color{#3D99F6}{k = 1} & n_1 = 1 = 2(\color{#3D99F6}{1}) - 1 \\ 2+3+4 & = 1^3 + \color{#3D99F6}{2}^3 & \color{#3D99F6}{k = 2} & n_2 = 3 = 2(\color{#3D99F6}{2}) - 1 \\ 5+6+7+8+9 & = 2^3 + \color{#3D99F6}{3}^3 & \color{#3D99F6}{k = 3} & n_3 = 5 = 2(\color{#3D99F6}{3}) - 1 \\ 10+11+12+13+14+15+16 & = 3^3 + \color{#3D99F6}{4}^3 & \color{#3D99F6}{k = 4} & n_4 = 7 = 2(\color{#3D99F6}{4}) - 1 \\ ... & ... & ... & ... \end{array}$

$\Rightarrow n_k = 2k - 1$ when $k = A+1$ , we have:

$\begin{aligned} \Rightarrow n_{A+1} & = 2(A+1) -1 = 10000 - 9802 + 1 = 199 \\ \Rightarrow A & = \boxed{99} \end{aligned}$

These proofs you guys are writing are very interesting. How would one write a proof for the number of terms on the LHS? I know that 2A+1 is the number of terms on the LHS. How would one write a proof?