Let Sn-k-1nk-k-14-3-k2-12-3-k-14-3 then which of the following is-are true-

The correct options are A S_7 =3 + 7^2/3/4 C S_26 =8 + 26^2/3/4Let t_k = k/(k 1)^4/3+(k 1)^2/3(k+1)^2/3+(k+1)^4/3 multiply N^r and D^r by (k+1^2/3 (k 1)^2/3) ...

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Byju's Answer

Standard XII

Mathematics

Method of Difference

Let Sn=∑k=1nk...

Question

Let $S_{n} = n \sum k = 1 \frac{k}{(k - 1)^{\frac{4}{3}} + (k^{2} - 1)^{\frac{2}{3}} + (k + 1)^{\frac{4}{3}}}$ then which of the following is/are true?

S_{7} = \frac{3 + 7^{\frac{2}{3}}}{4}

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S_{7} = \frac{7^{\frac{2}{3}} + 6^{\frac{2}{3}} - 1}{4}

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S_{26} = \frac{8 + 26^{\frac{2}{3}}}{4}

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S_{26} = \frac{26^{\frac{2}{3}} + 25^{\frac{2}{3}} - 1}{4}

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Solution

The correct options are
A $S_{7} = \frac{3 + 7^{\frac{2}{3}}}{4}$
C $S_{26} = \frac{8 + 26^{\frac{2}{3}}}{4}$
Let $t_{k} = \frac{k}{(k - 1)^{\frac{4}{3}} + (k - 1)^{\frac{2}{3}} (k + 1)^{\frac{2}{3}} + (k + 1)^{\frac{4}{3}}}$

multiply $N^{r} and D^{r} by (k + 1^{\frac{2}{3}} - (k - 1)^{\frac{2}{3}})$

$t_{k} = \frac{k ({(k + 1)}^{\frac{2}{3}} - (k - 1)^{\frac{2}{3}})}{4 k} = \frac{(k + 1)^{\frac{2}{3}} - (k - 1)^{\frac{2}{3}}}{4}$

So, $S_{n} = \frac{(n + 1)^{\frac{2}{3}} + n^{\frac{2}{3}} - 1}{4}$

So $S_{7} = \frac{3 + 7^{\frac{2}{3}}}{4}, S_{26} = \frac{8 + 26^{\frac{2}{3}}}{4}$

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Similar questions

Q. Let

S_{n} = n \sum k = 1 \frac{k}{(k - 1)^{\frac{4}{3}} + (k^{2} - 1)^{\frac{2}{3}} + (k + 1)^{\frac{4}{3}}}

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Q. Let

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S_{n} = \sum_{k = 1}^{n} \frac{n}{n^{2} + k n + k^{2}} a n d T_{n} = \sum_{k = 0}^{n - 1} \frac{n}{n^{2} + k n + k^{2}}

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Let Sn=n∑k=1k(k−1)43+(k2−1)23+(k+1)43 then which of the following is/are true?

The correct options are A S7=3+7234 C S26=8+26234Let tk=k(k−1)43+(k−1)23(k+1)23+(k+1)43 multiply Nr and Dr by (k+123−(k−1)23) tk=k((k+1)23−(k−1)23)4k=(k+1)23−(k−1)234 So, Sn=(n+1)23+n23−14 So S7=3+7234, S26=8+26234

Let $S_{n} = n \sum k = 1 \frac{k}{(k - 1)^{\frac{4}{3}} + (k^{2} - 1)^{\frac{2}{3}} + (k + 1)^{\frac{4}{3}}}$ then which of the following is/are true?