12-1 - 12 - 22-1 - 2 - 12 - 22 - 32-1 - 2 - 3 - - upto n terms is

The correct option is A 1/3 n(n + 2) General term of the given series is, T_n =∑_k=1^n k^2/∑_k=1^n k=16n(n+1)(2n+1)/12n(n+1)=1/3(2n+1) The required s ...

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Byju's Answer

Standard XII

Mathematics

Summation by Sigma Method

12/1 + 12 + 2...

Question

$\frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + . . . . .$ upto n terms is

\frac{1}{P} 3 (2 n + 1)

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\frac{1}{3} n^{2}

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\frac{1}{3} (n + 2)

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\frac{1}{3} n (n + 2)

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Solution

The correct option is A $\frac{1}{3} n (n + 2)$
General term of the given series is,
$T_{n} = \frac{\sum_{k = 1}^{n} k^{2}}{\sum_{k = 1}^{n} k} = \frac{\frac{1}{6} n (n + 1) (2 n + 1)}{\frac{1}{2} n (n + 1)} = \frac{1}{3} (2 n + 1)$
The required summation is,
$n \sum k = 1 T_{k} = \frac{1}{3} (2 n \sum k = 1 k + n \sum k = 1 1) = \frac{1}{3} [n (n + 1) + n] = \frac{1}{3} n (n + 2)$
Hence option 'D' is correct choice.

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Similar questions

Q. The sum of series

\frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + . . . .

upto n terms is

\frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + \dots + n

terms

=

Q. Show that

\frac{1 \times 2^{2} + 2 \times 3^{2} + . . . + n \times {(n + 1)}^{2}}{1^{2} \times 2 + 2^{2} \times 3 + . . . + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}

Show that

$\frac{1 \times 2^{2} + 2 \times 3^{2} + \dots \dots + n \times (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + \dots \dots + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}$

Find $(3^{3} - 2^{3}) + (5^{3} - 4^{3}) + (7^{3} - 6^{3}) + . . . .$ to 10 terms.

Show that $\frac{1 \times 2^{2} + 2 \times 3^{2} + . . . . + n \times (n + 1)^{2}}{1^{2} \times 2 + 2^{2} \times 3 + . . . . + n^{2} \times (n + 1)} = \frac{3 n + 5}{3 n + 1}$

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121+12+221+2+12+22+321+2+3+..... upto n terms is

The correct option is A 13n(n+2)General term of the given series is, Tn=∑nk=1k2∑nk=1k=16n(n+1)(2n+1)12n(n+1)=13(2n+1)The required summation is,n∑k=1Tk=13(2n∑k=1k+n∑k=11)=13[n(n+1)+n]=13n(n+2)Hence option 'D' is correct choice.

$\frac{1^{2}}{1} + \frac{1^{2} + 2^{2}}{1 + 2} + \frac{1^{2} + 2^{2} + 3^{2}}{1 + 2 + 3} + . . . . .$ upto n terms is