Animated Solution for Mathematics - Sequence and Series: Let $\alpha =1^2+4^2+8^2+13^2+19^2+26^2+\dots$ upto 10 terms and $\beta ={\sum }_{n=1}^{10}{n}^4$. If $4\alpha -\beta =55k+40$, then $k$ is equal to

To find the soul of this sequence, we must look at how it changes. Let us calculate the differences between consecutive terms:

Do you see it? The differences are $3,4,5,6,7,\dots$. This is an Arithmetic Progression!

When the first differences of a sequence form an A.P., the sequence itself must be a quadratic polynomial of the form $x_n=a{n}^2+bn+c$. This is a powerful realization that transforms a daunting guessing game into a structured algebraic problem.

Now, we need to find the specific values of $a$, $b$, and $c$. We have the first three terms of our sequence, so let us set up our equations:

For $n=1$: $a(1{)}^2+b(1)+c=1⟹a+b+c=1$

For $n=2$: $a(2{)}^2+b(2)+c=4⟹4a+2b+c=4$

For $n=3$: $a(3{)}^2+b(3)+c=8⟹9a+3b+c=8$

Solving this system is a rite of passage. Subtracting the first from the second gives $3a+b=3$. Subtracting the second from the third gives $5a+b=4$.

Subtracting these two results gives $2a=1$, so $a=\frac{1}{2}$. Substituting back, we find $b=\frac{3}{2}$ and $c=-1$. Thus, our general term is:

Now, we turn our attention to $\alpha ={\sum }_{n=1}^{10}{x}_n^2$. Substituting our general term, we get:

Expanding a trinomial square $(a+b+c{)}^2=a^2+b^2+c^2+2ab+2bc+2ca$ is a test of patience. Let us do it carefully:

We are asked to find $4\alpha -\beta$, where $\beta ={\sum }_{n=1}^{10}{n}^4$. Look at the elegance of this request!

When we multiply $\alpha$ by $4$, the denominator vanishes, and we are left with the sum of our expanded polynomial. When we subtract $\beta$, the $n^4$ terms vanish into thin air:

We use the standard summation formulas for the first $10$ natural numbers:

Plugging these in:

Finally, we equate this to $55k+40=19455$, which gives $55k=19415$, and thus $k=353$. You have navigated the sequence, conquered the expansion, and witnessed the beauty of algebraic cancellation.