Q.Find the 20th term of the series $2 \times 4 + 4 \times 6 + 6 \times 8 + \ldots + n$ terms.

We are asked to find $n$ such that

Concept and intuition: Sum of Products and Binomial Expectation

Each term is of the form $(k+1) \cdot \frac{{}^nC_k}{2^n}$. Notice that $\frac{{}^nC_k}{2^n}$ is exactly the probability that a binomial random variable $X$ with $n$ trials and success probability $1/2$ equals $k$. So the sum is

For a binomial distribution, $E[X] = n \cdot \frac12 = \frac{n}{2}$. Hence $S = \frac{n}{2} + 1$. Setting this equal to $16$ gives $\frac{n}{2} = 15$, so $n=30$.

Now let’s verify step by step without relying on probability, using algebra.

1. Write the sum in sigma notation

The factor $2^n$ is constant, so

2. Split the sum

The second sum is the total number of subsets of an $n$-element set: $\sum_{k=0}^n \binom{n}{k} = 2^n$.

3. Evaluate $\sum k \binom{n}{k}$ Use the identity $k \binom{n}{k} = n \binom{n-1}{k-1}$ (valid for $k \ge 1$). Then $\sum_{k=0}^n k \binom{n}{k} = \sum_{k=1}^n n \binom{n-1}{k-1} = n \sum_{j=0}^{n-1} \binom{n-1}{j} = n \cdot 2^{n-1}.$ …

Identify the series. The terms $2,6,16,40,\dots$ are

so the $r$-th term is $(r+1)2^{r-1}$, matching the Reason's series $2+3\cdot2+4\cdot2^2+\cdots$.

Verify the Reason.

So $2+3\cdot2+4\cdot2^2+\cdots$ to $n$ terms $=n\cdot 2^{n}$. R is true.

Verify the Assertion. Using this result, the sum to $k$ terms equals $k\cdot 2^{k}$. Set it to $4608$: …

Concept and Intuition

The two interleaved factor sequences $2,5,8,11,\dots$ and $5,9,13,17,\dots$ are both arithmetic progressions, so the $r$-th term of the product series is a simple quadratic in $r$; summing a quadratic-in-$r$ series over $r=1$ to $n$ always gives a cubic polynomial in $n$ (via the standard $\sum r^2,\sum r,\sum 1$ formulas), matching the given form $an^3+bn^2+cn+d$.

Step-by-Step Solution

1. First factors: $2,5,8,11,\dots \Rightarrow$ $r$-th term $=3r-1$.
2. Second factors: $5,9,13,17,\dots \Rightarrow$ $r$-th term $=4r+1$.
3. General term: $T_r=(3r-1)(4r+1)=12r^2+3r-4r-1=12r^2-r-1$.
4. Sum to $n$ terms: $S_n=12\sum r^2-\sum r-\sum 1 = 12\cdot\dfrac{n(n+1)(2n+1)}{6}-\dfrac{n(n+1)}{2}-n$.
5. Simplify: $12\cdot\dfrac{n(n+1)(2n+1)}{6}=2n(n+1)(2n+1)=4n^3+6n^2+2n$.
6. $S_n = 4n^3+6n^2+2n-\dfrac{n^2+n}{2}-n = \dfrac{8n^3+12n^2+4n-n^2-n-2n}{2}=\dfrac{8n^3+11n^2+n}{2}$.
7. So $a=4,\ b=\dfrac{11}{2},\ c=\dfrac12,\ d=0$. …

Concept and Intuition

The cube roots of unity satisfy $\omega^3=1$ and $1+\omega+\omega^2=0$, so $\omega+\omega^2=-1$ and $\omega\cdot\omega^2=\omega^3=1$. These two identities let any expression of the form $(n+\omega)(n+\omega^2)$ collapse to a clean polynomial in $n$: expanding gives $n^2+n(\omega+\omega^2)+\omega\cdot\omega^2 = n^2-n+1$. Recognising $1/\omega=\omega^2$ and $1/\omega^2=\omega$ (since $\omega\cdot\omega^2=1$) converts the given series into exactly this shape, after which the whole sum reduces to a routine sum of cubes/squares/naturals.

Step-by-Step Solution

1. Note $1/\omega=\omega^2$ and $1/\omega^2=\omega$ (because $\omega\cdot\omega^2=\omega^3=1$).
2. The $r$-th term of the series is $r\left((r+1)+\dfrac1\omega\right)\left((r+1)+\dfrac1{\omega^2}\right) = r\big((r+1)+\omega^2\big)\big((r+1)+\omega\big)$.
3. Let $n=r+1$. Expand $(n+\omega^2)(n+\omega) = n^2 + n(\omega+\omega^2) + \omega^3 = n^2 - n + 1$ (since $\omega+\omega^2=-1$, $\omega^3=1$).
4. Substitute back $n=r+1$: $(r+1)^2-(r+1)+1 = r^2+2r+1-r-1+1 = r^2+r+1$. …

Concept and Intuition

For each quadratic $x^2-(\text{sum})x+(\text{product})=0$, Vieta's formulas relate the roots' sum and product to the coefficients. Here we have two "coupled" quadratics — the coefficients of one involve the roots of the other — so writing down all four Vieta relations and eliminating variables lets us pin down the total sum $\alpha+\beta+\gamma+\delta$.

Step-by-Step Solution

1. $\alpha,\beta$ are roots of $x^2-5\gamma x-6\delta=0$, so by Vieta: $\alpha+\beta=5\gamma$ …(1), $\alpha\beta=-6\delta$ …(3).
2. $\gamma,\delta$ are roots of $x^2-5\alpha x-6\beta=0$, so by Vieta: $\gamma+\delta=5\alpha$ …(2), $\gamma\delta=-6\beta$ …(4).
3. The quantity sought, $S=\alpha+\beta+\gamma+\delta = (\alpha+\beta)+(\gamma+\delta) = 5\gamma+5\alpha = 5(\alpha+\gamma)$, so we just need $\alpha+\gamma$.
4. From (1): $\beta = 5\gamma-\alpha$. From (2): $\delta=5\alpha-\gamma$.
5. Substitute into (3): $\alpha(5\gamma-\alpha) = -6(5\alpha-\gamma) \Rightarrow 5\alpha\gamma-\alpha^2 = -30\alpha+6\gamma \Rightarrow 5\alpha\gamma-\alpha^2+30\alpha-6\gamma=0$. …(A)
6. Substitute into (4): $\gamma(5\alpha-\gamma) = -6(5\gamma-\alpha) \Rightarrow 5\alpha\gamma-\gamma^2 = -30\gamma+6\alpha \Rightarrow 5\alpha\gamma-\gamma^2+30\gamma-6\alpha=0$. …(B)
7. Subtract (B) from (A): $(-\alpha^2+30\alpha-6\gamma)-(-\gamma^2+30\gamma-6\alpha) = 0 \Rightarrow \gamma^2-\alpha^2+36\alpha-36\gamma=0$.
8. Factor: $(\gamma-\alpha)(\gamma+\alpha) - 36(\gamma-\alpha) = 0 \Rightarrow (\gamma-\alpha)(\gamma+\alpha-36)=0$. …

Concept and Intuition

The given series $2\cdot5 + 5\cdot9 + 8\cdot13 + 11\cdot17+\cdots$ pairs terms from two separate arithmetic progressions:

• First factor: $2,5,8,11,\ldots$ — AP with first term $2$, common difference $3$, so $n$-th term $=2+3(n-1)=3n-1$.
• Second factor: $5,9,13,17,\ldots$ — AP with first term $5$, common difference $4$, so $n$-th term $=5+4(n-1)=4n+1$.

The general term of the series is the product $T_n=(3n-1)(4n+1)$, a quadratic in $n$, so the sum can be evaluated using the standard formulas $\sum_{n=1}^{N}n = \dfrac{N(N+1)}2$ and $\sum_{n=1}^N n^2=\dfrac{N(N+1)(2N+1)}6$.

Step-by-Step Solution

1. Confirm the pattern: $T_1=2\cdot5=10$, $T_2=5\cdot9=45$, $T_3=8\cdot13=104$, $T_4=11\cdot17=187$ — matches $(3n-1)(4n+1)$ for $n=1,2,3,4$.
2. Expand: $T_n=(3n-1)(4n+1)=12n^2+3n-4n-1=12n^2-n-1$. …

Concept and Intuition

Whenever consecutive terms' denominators share a common difference (here, factors $3,7,11,15,\ldots$ each differing by $4$), splitting each term via partial fractions turns the sum into a telescoping one, where interior terms cancel and only the first and last survive.

Step-by-Step Solution

1. The $k$-th term (for $k=1,2,\ldots$) is $\dfrac{1}{(4k-1)(4k+3)}$: check $k=1$ gives $\dfrac{1}{3\cdot7}$, $k=2$ gives $\dfrac{1}{7\cdot11}$, etc. — matches the pattern.
2. Partial fractions: $\dfrac{1}{(4k-1)(4k+3)} = \dfrac14\left(\dfrac{1}{4k-1}-\dfrac{1}{4k+3}\right)$, since the difference between the two denominators is $4$.
3. Summing from $k=1$ to $50$:

Every interior term cancels (telescopes), leaving only the very first and very last pieces: …

Concept and Intuition

When binomial coefficients $\binom ni$ are weighted by a linear function of $i$ (here $3+5i$), split the sum into the two standard identities: $\sum_i\binom ni=2^n$ and $\sum_i i\binom ni = n\cdot2^{n-1}$ (the second follows from $i\binom ni = n\binom{n-1}{i-1}$).

Step-by-Step Solution

1. Note the coefficients $3,8,13,18,23,28$ increase by $5$ each time, so the $i$-th term's coefficient (for $\binom5i$, $i=0,\ldots,5$) is $3+5i$.
2. Sum $=\sum_{i=0}^{5}(3+5i)\binom5i = 3\sum_{i=0}^5\binom5i + 5\sum_{i=0}^5 i\binom5i$.
3. $\sum_{i=0}^5\binom5i = 2^5=32$. …

Concept and Intuition

For a quadratic $ax^2+bx+c=0$ with roots $r_1,r_2$: $r_1+r_2=-b/a$, $r_1r_2=c/a$. Power sums of roots can be built up algebraically: $r_1^2+r_2^2=(r_1+r_2)^2-2r_1r_2$, and then $r_1^4+r_2^4=(r_1^2+r_2^2)^2-2(r_1r_2)^2$ — squaring twice rather than solving for the (irrational) roots directly.

Step-by-Step Solution

1. For $16x^2-10x+1=0$: sum $S=r_1+r_2=10/16=5/8$; product $P=r_1r_2=1/16$.
2. $r_1^2+r_2^2 = S^2-2P = \dfrac{25}{64}-\dfrac{2}{16} = \dfrac{25}{64}-\dfrac{8}{64}=\dfrac{17}{64}$.
3. $r_1^4+r_2^4 = (r_1^2+r_2^2)^2 - 2(r_1r_2)^2 = \left(\dfrac{17}{64}\right)^2 - 2\left(\dfrac{1}{16}\right)^2$. …

Concept and Intuition

A product of cosines whose angles double each time ($\theta, 2\theta, 4\theta, \dots$) telescopes beautifully if you multiply and divide by $\sin\theta$, because $2\sin\theta\cos\theta=\sin2\theta$ collapses one factor at a time until only $\sin(2^n\theta)$ survives on top.

Step-by-Step Solution

1. Reorder the given product from smallest to largest angle: $\cos\frac{\pi}{32}\cos\frac{\pi}{16}\cos\frac{\pi}{8}\cos\frac{\pi}{4}$ (multiplication is commutative, so this is the same product).
2. Let $\theta=\pi/32$. Then the angles are $\theta,2\theta,4\theta,8\theta$, i.e. a 4-term double-angle chain.
3. Multiply and divide by $\sin\theta$: $\sin\theta\cos\theta = \tfrac12\sin2\theta$, so $\sin\theta\cos\theta\cos2\theta\cos4\theta\cos8\theta = \tfrac12\sin2\theta\cos2\theta\cos4\theta\cos8\theta = \tfrac14\sin4\theta\cos4\theta\cos8\theta=\tfrac18\sin8\theta\cos8\theta=\tfrac1{16}\sin16\theta.$ …

Concept and Intuition

Since $\alpha,\beta$ are roots of $x^2+2x+4=0$, Vieta's formulas give their sum and product directly, and standard symmetric-function identities let us get $\alpha^3+\beta^3$ (and hence the reciprocal sum) without solving for the (complex) roots themselves.

Step-by-Step Solution

1. From $x^2+2x+4=0$: $\alpha+\beta=-2$, $\alpha\beta=4$.
2. $\alpha^3+\beta^3 = (\alpha+\beta)^3 - 3\alpha\beta(\alpha+\beta) = (-2)^3 - 3(4)(-2) = -8+24 = 16$. …