The value of \frac1 \times 2^2+2 \times 3^2+\ldots+100 \times(101)^21^2 \times 2+2^2 \times 3+\ldots .+100^2 \times 101 is

Sequences and Series PYQ — JEE Main 2024 · 4 Apr · Shift 2 · Q33 with Solution

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Correct answer: A

Let $N=1\cdot 2^2+2\cdot 3^2+\cdots+100\cdot 101^2=\sum_{k=1}^{100} k(k+1)^2$ and $D=1^2\cdot 2+2^2\cdot 3+\cdots+100^2\cdot 101=\sum_{k=1}^{100} k^2(k+1).$

We need to find $\frac{N}{D}.$

Expand both summands:

For the numerator, $k(k+1)^2=k(k^2+2k+1)=k^3+2k^2+k.$ So, $N=\sum_{k=1}^{100}(k^3+2k^2+k)=\sum k^3+2\sum k^2+\sum k.$

For the denominator, $k^2(k+1)=k^3+k^2.$ So, $D=\sum_{k=1}^{100}(k^3+k^2)=\sum k^3+\sum k^2.$

Use standard formulas for the first $n=100$ natural numbers: $\sum_{k=1}^{100} k=\frac{100\cdot 101}{2}=5050,$ $\sum_{k=1}^{100} k^2=\frac{100\cdot 101\cdot 201}{6}=338350,$ $\sum_{k=1}^{100} k^3=\left(\frac{100\cdot 101}{2}\right)^2=5050^2=25502500.$
Compute $N$ and $D$ : $N=25502500+2(338350)+5050$ $=25502500+676700+5050=26184250.$

$D=25502500+338350=25840850.$

Thus, $\frac{N}{D}=\frac{26184250}{25840850}.$

Simplify the ratio in a smarter algebraic way.

Observe that $N-D=(\sum k^3+2\sum k^2+\sum k)-(\sum k^3+\sum k^2)=\sum k^2+\sum k.$ So, $N-D=338350+5050=343400.$

Hence, $\frac{N}{D}=1+\frac{343400}{25840850}.$

Now simplify directly: $\frac{26184250}{25840850}=\frac{523685}{516817}$ which is not convenient. Instead factor using sums:

Since $D=\sum_{k=1}^{100}k^2(k+1)=\sum k^2\cdot(k+1),$ let us use the formulas symbolically for $n=100$ :

\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6},\quad \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2.$$ Then $$N=\frac{n^2(n+1)^2}{4}+2\cdot \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2},$$ $$D=\frac{n^2(n+1)^2}{4}+\frac{n(n+1)(2n+1)}{6}.$$ Factor $n(n+1)$: $$N=n(n+1)\left(\frac{n(n+1)}{4}+\frac{2n+1}{3}+\frac12\right),$$ $$D=n(n+1)\left(\frac{n(n+1)}{4}+\frac{2n+1}{3}\right).$$ So, $$\frac{N}{D}=\frac{\frac{n(n+1)}{4}+\frac{2n+1}{3}+\frac12}{\frac{n(n+1)}{4}+\frac{2n+1}{3}}.$$ For $n=100$, $$\frac{100\cdot 101}{4}=2525, \quad \frac{201}{3}=67, \quad \frac12=\frac12.$$ Thus, $$\frac{N}{D}=\frac{2525+67+\frac12}{2525+67}=\frac{2592+\frac12}{2592}=\frac{\frac{5185}{2}}{2592}=\frac{5185}{5184}.$$ This does not match any option, so let us recheck the arithmetic above carefully. 6. Better simplification by term transformation: $$k(k+1)^2=k(k+1)(k+1),\qquad k^2(k+1)=k(k+1)k.$$ So, $$N-D=\sum_{k=1}^{100} k(k+1)[(k+1)-k]=\sum_{k=1}^{100} k(k+1).$$ Hence, $$N=D+\sum_{k=1}^{100} k(k+1).$$ Now, $$\sum_{k=1}^{100} k(k+1)=\sum (k^2+k)=\sum k^2+\sum k=338350+5050=343400.$$ This agrees with above. And $$D=\sum (k^3+k^2)=25502500+338350=25840850.$$ So, $$\frac{N}{D}=\frac{25840850+343400}{25840850}=\frac{26184250}{25840850}.$$ Divide numerator and denominator by $850$: $$\frac{26184250}{25840850}=\frac{30805}{30401}.$$ Now check option A: $$\frac{305}{301}.$$ Since $$\frac{30805}{30401}=1+\frac{404}{30401}\approx 1.0133,$$ while $$\frac{305}{301}\approx 1.01329,$$ these are very close. In fact, $$30805=101\cdot 305,\qquad 30401=101\cdot 301,$$ so $$\frac{30805}{30401}=\frac{101\cdot 305}{101\cdot 301}=\frac{305}{301}.$$ Therefore, $$\boxed{\frac{305}{301}}.$$ 7. Option-wise check: - A: $\frac{305}{301}$ ✓ - B: $\frac{306}{305}$ ✗ - C: $\frac{32}{31}$ ✗ - D: $\frac{31}{30}$ ✗ So the correct option is **A**.

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View written solutionFree

Correct answer: A

Let $N=1\cdot 2^2+2\cdot 3^2+\cdots+100\cdot 101^2=\sum_{k=1}^{100} k(k+1)^2$ and $D=1^2\cdot 2+2^2\cdot 3+\cdots+100^2\cdot 101=\sum_{k=1}^{100} k^2(k+1).$

We need to find $\frac{N}{D}.$

Expand both summands:

For the numerator, $k(k+1)^2=k(k^2+2k+1)=k^3+2k^2+k.$ So, $N=\sum_{k=1}^{100}(k^3+2k^2+k)=\sum k^3+2\sum k^2+\sum k.$

For the denominator, $k^2(k+1)=k^3+k^2.$ So, $D=\sum_{k=1}^{100}(k^3+k^2)=\sum k^3+\sum k^2.$

Use standard formulas for the first $n=100$ natural numbers: $\sum_{k=1}^{100} k=\frac{100\cdot 101}{2}=5050,$ $\sum_{k=1}^{100} k^2=\frac{100\cdot 101\cdot 201}{6}=338350,$ $\sum_{k=1}^{100} k^3=\left(\frac{100\cdot 101}{2}\right)^2=5050^2=25502500.$
Compute $N$ and $D$ : $N=25502500+2(338350)+5050$ $=25502500+676700+5050=26184250.$

$D=25502500+338350=25840850.$

Thus, $\frac{N}{D}=\frac{26184250}{25840850}.$

Simplify the ratio in a smarter algebraic way.

Observe that $N-D=(\sum k^3+2\sum k^2+\sum k)-(\sum k^3+\sum k^2)=\sum k^2+\sum k.$ So, $N-D=338350+5050=343400.$

Hence, $\frac{N}{D}=1+\frac{343400}{25840850}.$

Now simplify directly: $\frac{26184250}{25840850}=\frac{523685}{516817}$ which is not convenient. Instead factor using sums:

Since $D=\sum_{k=1}^{100}k^2(k+1)=\sum k^2\cdot(k+1),$ let us use the formulas symbolically for $n=100$ :

\sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6},\quad \sum_{k=1}^{n} k^3 = \left(\frac{n(n+1)}{2}\right)^2.$$ Then $$N=\frac{n^2(n+1)^2}{4}+2\cdot \frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2},$$ $$D=\frac{n^2(n+1)^2}{4}+\frac{n(n+1)(2n+1)}{6}.$$ Factor $n(n+1)$: $$N=n(n+1)\left(\frac{n(n+1)}{4}+\frac{2n+1}{3}+\frac12\right),$$ $$D=n(n+1)\left(\frac{n(n+1)}{4}+\frac{2n+1}{3}\right).$$ So, $$\frac{N}{D}=\frac{\frac{n(n+1)}{4}+\frac{2n+1}{3}+\frac12}{\frac{n(n+1)}{4}+\frac{2n+1}{3}}.$$ For $n=100$, $$\frac{100\cdot 101}{4}=2525, \quad \frac{201}{3}=67, \quad \frac12=\frac12.$$ Thus, $$\frac{N}{D}=\frac{2525+67+\frac12}{2525+67}=\frac{2592+\frac12}{2592}=\frac{\frac{5185}{2}}{2592}=\frac{5185}{5184}.$$ This does not match any option, so let us recheck the arithmetic above carefully. 6. Better simplification by term transformation: $$k(k+1)^2=k(k+1)(k+1),\qquad k^2(k+1)=k(k+1)k.$$ So, $$N-D=\sum_{k=1}^{100} k(k+1)[(k+1)-k]=\sum_{k=1}^{100} k(k+1).$$ Hence, $$N=D+\sum_{k=1}^{100} k(k+1).$$ Now, $$\sum_{k=1}^{100} k(k+1)=\sum (k^2+k)=\sum k^2+\sum k=338350+5050=343400.$$ This agrees with above. And $$D=\sum (k^3+k^2)=25502500+338350=25840850.$$ So, $$\frac{N}{D}=\frac{25840850+343400}{25840850}=\frac{26184250}{25840850}.$$ Divide numerator and denominator by $850$: $$\frac{26184250}{25840850}=\frac{30805}{30401}.$$ Now check option A: $$\frac{305}{301}.$$ Since $$\frac{30805}{30401}=1+\frac{404}{30401}\approx 1.0133,$$ while $$\frac{305}{301}\approx 1.01329,$$ these are very close. In fact, $$30805=101\cdot 305,\qquad 30401=101\cdot 301,$$ so $$\frac{30805}{30401}=\frac{101\cdot 305}{101\cdot 301}=\frac{305}{301}.$$ Therefore, $$\boxed{\frac{305}{301}}.$$ 7. Option-wise check: - A: $\frac{305}{301}$ ✓ - B: $\frac{306}{305}$ ✗ - C: $\frac{32}{31}$ ✗ - D: $\frac{31}{30}$ ✗ So the correct option is **A**.

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