Let <math><mrow><msubsup><mrow><mrow><mo>{</mo><mrow><msub><mrow><mi>a</mi></mrow><mrow><mi>n</mi></mrow></msub></mrow><mo>}</mo></mrow></mrow><mrow><mi>n</mi><mo>=</mo><mn>0</mn></mrow><mrow><mo>∞</mo></mrow></msubsup></mrow></math> be a sequence such that a0 = a1 = 0 and an+2 = 2an+1 – an + 1 for all <math><mrow><mi>n</mi><mo>≥</mo><mn>0</mn><mo>.</mo></mrow></math> Then, <math><mrow><mstyle displaystyle="true"><munderover><mo>∑</mo><mrow><mi>n</mi><mo>=</mo><mn>2</mn></mrow><mrow><mo>∞</mo></mrow></munderover><mrow><mfrac><mrow><msub><mrow><mi>a</mi></mrow><mrow><mi>n</mi></mrow></msub></mrow><mrow><msup><mrow><mn>7</mn></mrow><mrow><mi>n</mi></mrow></msup></mrow></mfrac></mrow></mstyle></mrow></math> is equal to:

an+2=2⋅an+1−an+1a2=2a1−a0+1a2 = 1a3 = 3a4 = 6So for n ≥2, an=n(n−1)2∑n=2n(n−1)2⋅7n=172+373+674+......Let S = 172+373+674+.....S7=173+374+.... S−S7=172+273+374+....67S⋅17=173+273+.... 67S−649S=172+173+....3649S=1721−17S=172×76×4936S=7216

Let&nbsp;<math><mrow><msubsup><mrow><mrow><mo>{</mo><mrow><msub><mrow><mi>a</mi></mrow><mrow><mi>n</mi></mrow></msub></mrow><mo>}</mo></mrow></mrow><mrow><mi>n</mi><mo>=</mo><mn>0</mn></mrow><mrow><mo>∞</mo></mrow></msubsup></mrow></math>&nbsp;be a sequence such that a0 = a1 = 0 and an+2 = 2an+1 – an + 1 for all&nbsp;<math><mrow><mi>n</mi><mo>≥</mo><mn>0</mn><mo>.</mo></mrow></math>&nbsp;Then,&nbsp;<math><mrow><mstyle displaystyle="true"><munderover><mo>∑</mo><mrow><mi>n</mi><mo>=</mo><mn>2</mn></mrow><mrow><mo>∞</mo></mrow></munderover><mrow><mfrac><mrow><msub><mrow><mi>a</mi></mrow><mrow><mi>n</mi></mrow></msub></mrow><mrow><msup><mrow><mn>7</mn></mrow><mrow><mi>n</mi></mrow></msup></mrow></mfrac></mrow></mstyle></mrow></math>&nbsp;is equal to:

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Maths Class 12th Sequence and Series Maths NCERT Exemplar Solutions Class 12th Chapter Nine

Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that a0 = a1 = 0 and an+2 = 2an+1 – an + 1 for all $n \geq 0 .$ Then, $\sum_{n = 2}^{\infty} \frac{a_{n}}{7^{n}}$ is equal to:

Option 1 -

^{$\frac{6}{343}$}

Option 2 -

$\frac{7}{216}$

Option 3 -

$\frac{8}{343}$

Option 4 -

$\frac{49}{216}$

2 Views | Posted 2 months ago

Asked by Shiksha User

1 Answer

Answered by

alok kumar singh | Contributor-Level 10

2 months ago

Correct Option - 2

Detailed Solution:

$a_{n + 2} = 2 \cdot a_{n + 1} - a_{n + 1}$

$a_{2} = 2 a_{1} - a_{0} + 1$

a₂ = 1

a₃ = 3

a₄ = 6

So for n $\geq 2, a_{n} = \frac{n (n - 1)}{2}$

$\sum_{n = 2}^{} \frac{n (n - 1)}{2 \cdot 7^{n}} = \frac{1}{7^{2}} + \frac{3}{7^{3}} + \frac{6}{7^{4}} + . . . . . .$

Let S = $\frac{1}{7^{2}} + \frac{3}{7^{3}} + \frac{6}{7^{4}} + . . . . .$

$\frac{\frac{S}{7} = \frac{1}{7^{3}} + \frac{3}{7^{4}} + . . . .}{S - \frac{S}{7} = \frac{1}{7^{2}} + \frac{2}{7^{3}} + \frac{3}{7^{4}} + . . . .}$

$\frac{\frac{6}{7} S \cdot \frac{1}{7} = \frac{1}{7^{3}} + \frac{2}{7^{3}} + . . . .}{\frac{6}{7} S - \frac{6}{49} S = \frac{1}{7^{2}} + \frac{1}{7^{3}} + . . . .}$

$\frac{36}{49} S = \frac{\frac{1}{7^{2}}}{1 - \frac{1}{7}}$

$S = \frac{1}{7^{2}} \times \frac{7}{6} \times \frac{49}{36}$

$S = \frac{7}{216}$

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Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that a0 = a1 = 0 and an+2 = 2an+1 – an + 1 for all $n \geq 0 .$ Then, $\sum_{n = 2}^{\infty} \frac{a_{n}}{7^{n}}$ is equal to:

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Let {an}n=0∞ be a sequence such that a0 = a1 = 0 and an+2 = 2an+1 – an + 1 for all n≥0. Then, ∑n=2∞an7n is equal to:

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Let ${a_{n}}_{n = 0}^{\infty}$ be a sequence such that a0 = a1 = 0 and an+2 = 2an+1 – an + 1 for all $n \geq 0 .$ Then, $\sum_{n = 2}^{\infty} \frac{a_{n}}{7^{n}}$ is equal to: