Find 33-23-53-43-73-63- - to 10 terms-OrShow that 1 - 22-2 - 32- -n -n-12-12- 2-22- 3-n2-n-1-3 n-5-3 n-1

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Byju's Answer

Standard VIII

Mathematics

Power of a Power

Find 33-23+53...

Question

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

Or

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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Solution

$S_{1} 0 = 4960$

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Similar questions

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

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

{lim}_{n \to \infty} \frac{1. n^{2} + 2 (n - 1)^{2} + 3 (n - 2)^{2} + . . . + n {.1}^{2}}{1^{3} + 2^{3} + 3^{3} + . . . + n^{3}}

1^{3} + 1^{2} + 1 + 2^{3} + 2^{2} + 2 + 3^{3} + 3^{2} + 3 + . . . 3 n

terms

=

L t_{n \to \infty} \frac{1. n^{2} + 2 (n - 1)^{2} + 3 (n . - 2)^{2} + \dots + n {.1}^{2}}{1^{3} + 2^{3} + . . + n^{3}}

equals to:

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Find (33−23)+(53−43)+(73−63)+.... to 10 terms. Or Show that 1×22+2×32+....+n×(n+1)212×2+22×3+....+n2×(n+1)=3n+53n+1

S10=4960

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

Or

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

$S_{1} 0 = 4960$